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Factorization of Quadratic Trinomials is the process of finding factors of given Quadratic Trinomials. If ax^2 + bx + c is an expression where a, b, c are constants, then the expression is called a quadratic trinomial in x. The expression ax^2 + bx + c has an x^2 term, x term, and an independent term. Find Factoring Quadratics Problems with Solutions in this article.

Factorization of Quadratic Trinomials Forms

The Factorization of Quadratic Trinomials is in two forms.

(i) First form: x^2 + px + q
(ii) Second form: ax^2 + bx + c

How to find Factorization of Trinomial of the Form x^2 + px + q?

If x^2 + px + q is an Quadratic Trinomial, then x² + (m + n) × + mn = (x + m)(x + n) is the identity.

Solved Examples on Factorization of Quadratic Trinomial of the Form x^2 + px + q 

1. Factorize the algebraic expression of the form x^2 + px + q

(i) a² – 7a + 12

Solution:
The Given expression is a² – 7a + 12.
By comparing the given expression a² – 7a + 12 with the basic expression x^2 + px + q.
Here, a = 1, b = – 7, and c = 12.
The sum of two numbers is m + n = b = – 7 = – 4 – 3.
The product of two number is m * n = a * c = -4 * (- 3) = 12
From the above two instructions, we can write the values of two numbers m and n as – 4 and -3.
Then, a² – 7a + 12 = a² – 4a -3a + 12.
= a (a – 4) – 3(a – 4).
Factor out the common terms.

Then, a² – 7a + 12 = (a – 4) (a – 3).

(ii) a² + 2a – 15

Solution:
The Given expression is a² + 2a – 15.
By comparing the given expression a² + 2a – 15 with the basic expression x^2 + px + q.
Here, a = 1, b = 2, and c = -15.
The sum of two numbers is m + n = b = 2 = 5 – 3.
The product of two number is m * n = a * c = 5 * (- 3) = -15
From the above two instructions, we can write the values of two numbers m and n as 5 and -3.
Then, a² + 2a – 15 = a² + 5a – 3a – 15.
= a (a + 5) – 3(a + 5).
Factor out the common terms.

Then, a² + 2a – 15 = (a + 5) (a – 3).

How to find Factorization of trinomial of the form ax^2 + bx + c?

To factorize the expression ax^2 + bx + c we have to find the two numbers p and q, such that p + q = b and p × q = ac

Solved Examples on Factorization of trinomial of the form ax^2 + bx + c 

2. Factorize the algebraic expression of the form ax2 + bx + c

(i) 15b² – 26b + 8

Solution:
The Given expression is 15b² – 26b + 8.
By comparing the given expression 15b² – 26b + 8 with the basic expression ax² + bx + c.
Here, a = 15, b = -26, and c = 8.
The sum of two numbers is p + q = b = -26 = 5 – 3.
The product of two number is p * q = a * c = 15 * (8) = 120
From the above two instructions, we can write the values of two numbers p and q as -20 and -6.
Then, 15b² – 26b + 8 = 15b² – 20 – 6b + 8.
= 5b (3b – 4) – 2(3b – 4).
Factor out the common terms.

Then, 15b² – 26b + 8 = (3b – 4) (5b – 2).

(ii) 3a² – a – 4

Solution:
The Given expression is 3a² – a – 4.
By comparing the given expression 3a² – a – 4 with the basic expression ax² + bx + c.
Here, a = 3, b = -1, and c = -4.
The sum of two numbers is p + q = b = -1 = 5 – 3.
The product of two number is p * q = a * c = 3 * (-4) = -12
From the above two instructions, we can write the values of two numbers p and q as -4 and 3.
Then, 3a² – a – 4 = 3a² – 4a -3a – 4.
= a (3a – 4) – 1(3a – 4).
Factor out the common terms.

Then, 3a² – a – 4 = (3a – 4) (a – 1).

Learn the process to Factorize the Trinomial ax Square Plus bx Plus c. One of the basic expressions for trinomial is ax² + bx + c. To find the ax² + bx + c factors, firstly, we need to find the two numbers and that is p and q. Here, the second term ‘b’ is the sum of the two numbers that is p + q = b. The product of the first and last terms is equal to the product of two numbers that is p * q = ac. Based on these two instructions, we need to find the values of p and q.

Steps to Factorize the Trinomial of Form ax² + bx + c?

1. Note down the given expression and compare it with the basic expression ax² + bx + c.
2. Note down the product and sum terms and find the two numbers.
3. Depends on the values of two numbers, expand the given expression.
4. Factor out the common terms.
5. Finally, we will get the product of two terms which is equal to the trinomial expression.

Solved Examples on Factoring Trinomials of the Form ax² + bx + c

1. Resolve into factors.

(i) 2s² + 9s + 10.

Solution:
The Given expression is 2s² + 9s + 10.
By comparing the given expression 2s² + 9s + 10 with the basic expression ax² + bx + c.
Here, a = 2, b = 9, and c = 10.
The sum of two numbers is p + q = b = 9 = 5 + 4.
The product of two number is p * q = a * c = 2 * 10 = 20 = 5 * 4.
From the above two instructions, we can write the values of two numbers p and q as 5 and 4.
Then, 2s² + 9s + 10 = 2s² + 5s + 4s + 20.
= 2s (s + 5) + 4 (s + 5).
Factor out the common terms.

Then, 2s² + 9s + 10 = (2s + 4) (s + 5).

(ii) 6s² + 7s –3

Solution:
The Given expression is 6s² + 7s – 3.
By comparing the given expression 6s² + 7s – 3 with the basic expression ax² + bx + c.
Here, a = 6, b = 7, and c = 3.
The sum of two numbers is p + q = b = 7 = 9 – 2.
The product of two number is p * q = a * c = 6 * 3 = 18 = 9 * 2.
From the above two instructions, we can write the values of two numbers p and q as 9 and 2.
Then, 6s² + 7s -3 = 6s² + 9s – 2s – 3.
= 6s² – 2s + 9s – 3.
= 2s (3s – 1) + 3(3s – 1).
Factor out the common terms.

Then, 6s² + 7s – 3 = (3s – 1) (2s + 3).

2. Factorize the trinomial.

(i) 2x² + 7x + 3.

Solution:
The Given expression is 2x² + 7x + 3.
By comparing the given expression 2x² + 7x + 3 with the basic expression ax² + bx + c.
Here, a = 2, b = 7, and c = 3.
The sum of two numbers is p + q = b = 7 = 6 + 1.
The product of two number is p * q = a * c = 2 * 3 = 6 = 6 * 1.
From the above two instructions, we can write the values of two numbers p and q as 6 and 1.
Then,2x² + 7x + 3 = 2x² + 6x + x + 3.
= 2x (x + 3) + (x + 3).
Factor out the common terms.

Then, 2x² + 7x + 3 = (x + 3) (2x + 1).

(ii) 3s² – 4s – 4.

Solution:
The Given expression is 3s² – 4s – 4.
By comparing the given expression 3s² – 4s – 4 with the basic expression ax² + bx + c.
Here, a = 3, b = – 4, and c = – 4.
The sum of two numbers is p + q = b = – 4 = – 6 + 2.
The product of two number is p * q = a * c = 3 * (- 4) = – 12 = (- 6) * 2.
From the above two instructions, we can write the values of two numbers p and q as – 6 and 2.
Then, 3s² – 4s – 4 = 3s² – 6s + 2s – 4.
= 3s (s – 2) + 2(s – 2).
Factor out the common terms.

Then, 3s² – 4s – 4 = (s – 2) (3s + 2).

Learn How to Factorize the Trinomial x² + px +q? A Trinomial is a three-term algebraic expression. By Factoring the trinomial expression, we will get the product of two binomial terms. Here, the trinomial expression contains three terms which are combined with the operations like addition or subtraction. Here, we need to find the coefficient values, and based on the coefficient values, we can find out the binomial terms as products of trinomial expression.

How to Factorize the Trinomial x² + px + q?

To find x^2 + px + q, we have to find the two terms (m + n) = p and mn = q.
Substitute (m + n) = p and mn = q in x^2 + px + q.
x^2 + px + q = x^2 + (m + n)x + mn.
By expanding the above expression, we will get
x^2 + px + q = x^2 + mx + nx + mn.
separate the common terms from the above expression.
that is, x(x + m) + n(x + m).
factor out the common term.
that is, (x + m) (x + n).
So, x^2 + px + q = (x + m)(x + n).

Factorization of Trinomial Steps

• Note down the given trinomial expression and compare the expression with the basic expression.
• Find out the product and sum of co-efficient values that is (m + n) and mn.
• Based on the above step, find out the two co-efficient values m and n.
• Finally, we will get the product of two terms which are equal to the trinomial expression.

Examples on Factoring Trinomials of Form x² + px + q

1. Resolve into factors

(i) a² + 3a -28

Solution:
Given Expression is a² + 3a -28.
Compare the a² + 3a -28 with the x^2 + px +q
Here, p = m + n = 3 and q = mn = -28
q is the product of two co-efficient. That is, 7 *(- 4) = -28
p is the sum of two co-efficient. That is 7 + ( – 4) = 3.
So, a² + 3a -28 = a² + [7 + (-4)]a – 28.
= a² + 7a – 4a – 28.
=a (a + 7) – 4(a + 7)
Factor out the common term.
That is, (a + 7) (a – 4).

Finally, the expression a² + 3a -28 = (a + 7) (a – 4).

(ii) a² + 8a + 15

Solution:
Given Expression is a² + 8a + 15.
Compare the a² + 8a + 15 with the x^2 + px +q.
Here, p = m + n = 8 and q = mn = 15.
q is the product of two co-efficient. That is, 5 * 3 = 15.
p is the sum of two co-efficient. That is 5 + 3 = 8.
So, a² + 8a + 15 = a² + (5 + 3)a + 15.
= a² + 5a + 3a + 15.
=a (a + 5) + 3(a + 5).
Factor out the common term.
That is, (a + 5) (a + 3).

Finally, the expression a² + 8a + 15= (a + 5) (a + 3).

2. Factorize the Trinomial

(i) a² + 15a + 56

Solution:
Given Expression is a² + 15a + 56.
Compare the a² + 15a + 56 with the x^2 + px +q.
Here, p = m + n = 15 and q = mn = 56.
q is the product of two co-efficient. That is, 7 * 8 = 56.
p is the sum of two co-efficient. That is 7 + 8 = 15.
So, a² + 15a + 56= a² + (7 + 8) a + 56.
= a² + 7a + 8a + 56.
=a (a + 7) + 8(a + 7).
Factor out the common term.
That is, (a + 7) (a + 8).

Finally, the expression a² + 15a + 56= (a + 7) (a + 8).

(ii) a² + a – 56

Solution:
Given Expression is a² + a – 56.
Compare the a² + a – 56 with the x^2 + px +q.
Here, p = m + n = 1 and q = mn = – 56.
q is the product of two co-efficient. That is, – 7 * 8 = – 56.
p is the sum of two co-efficient. That is – 7 + 8 = 1.
So, a² + a – 56 = a² + ( – 7 + 8)a – 56.
= a² – 7a + 8a – 56.
=a (a – 7) + 8(a – 7).
Factor out the common term.
That is, (a – 7) (a + 8).

Finally, the expression a^2 + a – 56 = (a – 7) (a + 8).

Are you looking for How to Evaluate the Difference of Two Squares Problems? We have given all Difference of Two Squares problems along with the evaluation of Difference of Two Squares with detailed explanation. Students can refer to all factorization problems on our website and begin their practice to score good marks in the exam.

Solved Problems to Evaluate the Difference of Two Squares

Use the formula of the difference of two squares to evaluate the following algebraic expressions:

(i) (202)² – (123)²

Solution:
Given expression is (202)² – (123)²
The above equation (202)² – (123)² is in the form of a² – b².
(202)² – (123)²
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 202 and b = 123
(202 + 123) (202 – 123)
(325) (79)
25675

The final answer is 25675.

(ii) (600)² – (598)²

Solution:
Given expression is (600)² – (598)²
The above equation (600)² – (598)² is in the form of a² – b².
(600)² – (598)²
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 600 and b = 598
(600 + 598) (600 – 598)
(1198) (2)
2396

The final answer is 2396.

(iii) (4.2)² – (2.1)²

Solution:
Given expression is (4.2)² – (2.1)²
The above equation (4.2)² – (2.1)² is in the form of a² – b².
(4.2)² – (2.1)²
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 4.2 and b = 2.1
(4.2 + 2.1) (4.2 – 2.1)
(6.3) (2.1)
13.23

The final answer is 13.23.

(iv) (97.8)² – (0.4)²

Solution:
Given expression is (97.8)² – (0.4)²
The above equation (97.8)² – (0.4)² is in the form of a² – b².
(97.8)² – (0.4)²
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 97.8 and b = 0.4
(97.8 + 0.4) (97.8 – 0.4)
(98.2) (97.4)
9564.68

The final answer is 9564.68.

(v) (8.4)² – (1.8)²

Solution:
Given expression is (8.4)² – (1.8)²
The above equation (8.4)² – (1.8)² is in the form of a² – b².
(8.4)² – (1.8)²
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 8.4 and b = 1.8
(8.4 + 1.8) (8.4 – 1.8)
(10.2) (6.6)
67.32

The final answer is 67.32.

Factoring a polynomial is the product of the two or more polynomials. Learn How to Factorize the Difference of Two Squares in this article. Break down all the huge algebraic expressions into small factors with the help of factorization. Solved Problems on Factoring the Difference of Two Squares are explained clearly along with the solutions. Visit all factorization problems and get complete knowledge of the factorization concept.

Solved Problems on How to Factorize the Difference of Two Squares

1. Factorize the following algebraic expressions

(i) m² – 121

Solution:
Given expression is m² – 121
Rewrite the above expression.
m² – (11)² 
The above equation m² – (11)² is in the form of a² – b².
m² – (11)² 
Now, apply the formula of a² – b² = (a + b) (a – b), where a = m and b = 11
(m + 11) (m – 11)

The final answer is (m + 11) (m – 11)

(ii) 49a² – 16b²

Solution:
Given expression is 49a² – 16b²
Rewrite the above expression.
(7a)² – (4b)² 
The above equation (7a)² – (4b)²  is in the form of a² – b².
(7a)² – (4b)² 
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 7a and b = 4b
(7a + 4b) (7a – 4b)

The final answer is (7a + 4b) (7a – 4b)

2. Factor the following

(i) 48m² – 243n²

Solution:
Given expression is 48m² – 243n²
Take 3 common
3{16m² – 81n²}
Rewrite the above expression.
3{(4m)² – (9n)²} 
The above equation {(4m)² – (9n)²}   is in the form of a² – b².
{(4m)² – (9n)²} 
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 4m and b = 9n
(4m + 9n) (4m – 9n)
3{(4m + 9n) (4m – 9n)}

The final answer is 3{(4m + 9n) (4m – 9n)}

(ii) 3a³ – 48a

Solution:
Given expression is 3a³ – 48a
Take 3 common
3a{a² – 16}
Rewrite the above expression.
3a{(a)² – (4)²} 
The above equation {(a)² – (4)²}    is in the form of a² – b².
{(a)² – (4)²} 
Now, apply the formula of a² – b² = (a + b) (a – b), where a = a and b = 4
(a + 4) (a – 4)
3a{(a + 4) (a – 4)}

The final answer is 3a{(a + 4) (a – 4)}

3. Factor the expressions

(i) 25(a + 3b)² – 16 (a – 3b)²

Solution:
Given expression is 25(a + 3b)² – 16 (a – 3b)²
Rewrite the above expression.
{[5(a + 3b)]² – [4 (a – 3b)]²} 
The above equation {[5(a + 3b)]² – [4 (a – 3b)]²} is in the form of a² – b².
{[5(a + 3b)]² – [4 (a – 3b)]²}
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 5(a + 3b) and b = 4 (a – 3b)
(5(a + 3b) + 4 (a – 3b)) (5(a + 3b) – [4 (a – 3b)])
(5a + 15b + 4a – 12b) (5a + 15b – 4a + 12b)
(9a + 3b) (a + 27b)
3(3a + b) (a + 27b)

The final answer is 3(3a + b) (a + 27b)

(ii) 4x² – 16/(25x²)

Solution:
Given expression is 4x² – 16/(25x²)
Rewrite the above expression.
{[2x]² – [4/5x]²}
The above equation {[2x]² – [4/5x]²} is in the form of a² – b².
{[2x]² – [4/5x]²}
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 2x and b = 4/5x
(2x + 4/5x) (2x – 4/5x)

The final answer is (2x + 4/5x) (2x – 4/5x)

Join the list of successful students who learned all the Factorization problems using the best material provided on our website. Yes, it’s true. Students can easily score good marks in the exam using our information on Factoring Differences of Squares. Also, we are giving all Factorization-Related Solved Problems, Extra Questions, and also Multiple Choice Questions and Answers. Utilize this opportunity and start your practice now and test your knowledge on the concept.

Factoring the Differences of Two Squares Examples

1. Factorize the following algebraic expressions

(i) 64 – a²

Solution:
Given expression is 64 – a²
Rewrite the above expression.
8² – a²
The above equation 8² – a² is in the form of a² – b².
[(8)² – (a)²]
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 8 and b = a
(8 + a) (8 – a)

The final answer is (8 + a) (8 – a)

(ii) 3m² – 27n²

Solution:
Given expression is 3m² – 27n²
Rewrite the above expression. Take 3 common.
3 (m² – (3n)²) where 9n² = (3n)²
The above equation (m² – (3n)²)  is in the form of a² – b².
[(m)² – (3n)²]
Now, apply the formula of a² – b² = (a + b) (a – b), where a = m and b = 3n
(m + 3n) (m – 3n)
3{(m + 3n) (m – 3n)}

The final answer is 3{(m + 3n) (m – 3n)}

(iii) a³ – 25a

Solution:
Given expression is a³ – 25a
Rewrite the above expression. Take a common.
a (a² – 25)
a ((a)² – (5)²)
The above equation ((a)² – (5)²) is in the form of a² – b².
((a)² – (5)²)
Now, apply the formula of a² – b² = (a + b) (a – b), where a = a and b = 5
(a + 5) (a – 5)
a {(a + 5) (a – 5)}

The final answer is a {(a + 5) (a – 5)}

2. Factor the expressions

(i) 81x² – (y – z)²

Solution:
Given expression is 81x² – (y – z)²
Rewrite the above expression.
(9x)² – (y – z)²
The above equation ((9x)² – (y – z)²) is in the form of a² – b².
((9x)² – (y – z)²)
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 9x and b = y – z
(9x + (y – z)) (9x – (y – z))
(9x + y – z) (9x – y + z)

The final answer is (9x + y – z) (9x – y + z)

(ii) 25(a + b)² – 36(a – 2b)².

Solution:
Given expression is 25(a + b)² – 36(a – 2b)²
Rewrite the above expression.
{5(a + b)}² – {6(a – 2b)}²
The above equation {5(a + b)}² – {6(a – 2b)}² is in the form of a² – b².
((5(a + b))² – (6(a – 2b))²)
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 5(a + b) and b = 6(a – 2b)
[5(a + b) + 6(a – 2b)] [5(a + b) – 6(a – 2b)]
[5a + 5b + 6a – 12b] [5a + 5b – 6a + 12b]
[11a – 7b] [17b – a]

The final answer is [11a – 7b] [17b – a]

(iii) (m – 2)² – (m – 3)²

Solution:
Given expression is (m – 2)² – (m – 3)²
The above equation (m – 2)² – (m – 3)² is in the form of a² – b².
(m – 2)² – (m – 3)²
Now, apply the formula of a² – b² = (a + b) (a – b), where a = m – 2 and b = m – 3
[(m – 2) + (m – 3)] [(m – 2) – (m – 3)]
[m – 2 + m – 3] [m – 2 – m + 3]
[2m – 5] [1]
[2m – 5]

The final answer is [2m – 5]

If you are searching for help on Factorization of Perfect Square Trinomials you are the correct place. Check out all the problems explained in this article. We have given a detailed explanation for every individual problem along with correct answers. Also, we are providing all factorization concepts, worksheets on our website for free of cost. Therefore, learn every concept and improve your knowledge easily.

Learn to solve the given algebraic expressions using the below formulas.
(i) a² + 2ab + b² = (a + b)² = (a + b) (a + b)
(ii) a² – 2ab + b² = (a – b)² = (a – b) (a – b)

Factoring Perfect Square Trinomials Examples

1. Factorization when the given expression is a perfect square

(i) m⁴ – 10m²n² + 25n⁴

Solution:
Given expression is m⁴ – 10m²n² + 25n⁴
The given expression m⁴ – 10m²n² + 25n⁴ is in the form a² – 2ab + b².
So find the factors of given expression using a² – 2ab + b² = (a – b)² = (a – b) (a – b) where a = m², b = 5n²
Apply the formula and substitute the a and b values.
m⁴ – 10m²n² + 25n⁴
(m²)² – 2 (m²) (5n²) + (5n²)²
(m² – 5n²)²
(m² – 5n²) (m² – 5n²)

Factors of the m⁴ – 10m²n² + 25n⁴ are (m² – 5n²) (m² – 5n²)

(ii) b²+ 6b + 9

Solution:
Given expression is b²+ 6b + 9
The given expression b²+ 6b + 9 is in the form a² + 2ab + b².
So find the factors of given expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = b, b = 3
Apply the formula and substitute the a and b values.
b²+ 6b + 9
(b)² + 2 (b) (3) + (3)²
(b + 3)²
(b + 3) (b + 3)

Factors of the b²+ 6b + 9 are (b + 3) (b + 3)

(iii) p⁴ – 2p² q² + q⁴

Solution:
Given expression is p⁴ – 2p² q² + q⁴
The given expression p⁴ – 2p² q² + q⁴ is in the form a² – 2ab + b².
So find the factors of given expression using a² – 2ab + b² = (a – b)² = (a – b) (a – b) where a = p², b = q²
Apply the formula and substitute the a and b values.
p⁴ – 2p² q² + q⁴
(p²)² – 2 (p²) (q²) + (q²)²
(p² – q²)²
(p² – q²) (p² – q²)
From the formula (a² – b²) = (a + b) (a – b), rewrite the above equation.
(p + q) (p – q) (p + q) (p – q)

Factors of the p⁴ – 2p² q² + q⁴ are (p + q) (p – q) (p + q) (p – q)

2. Factor using the identity

(i) 25 – a² – 2ab – b²

Solution:
Given expression is 25 – a² – 2ab – b²
Rearrange the given expression as 25 – (a² + 2ab + b²)
a² + 2ab + b² = (a + b)² = (a + b) (a + b)
25 – (a + b)²
(5)²– (a + b)²
From the formula (a² – b²) = (a + b) (a – b), rewrite the above equation.
[(5 + a + b)(5 – a – b)]

(ii) 1- 2mn – (m² + n²)

Solution:
Given expression is 1- 2mn – (m² + n²)
1- 2mn – m² – n²
1 – (2mn + m² + n²)
1 – (m + n)²
(1)² – (m + n)²
From the formula (a² – b²) = (a + b) (a – b), rewrite the above equation.
(1 + m + n) (1 – m + n)

Factorization of Perfect Square is the process of finding factors for an equation which is in the form of a² + 2ab + b² or a² – 2ab + b². Get to know the step by step procedure involved for finding factors of a perfect square. Have a look at the different examples taken to illustrate the Factorization of Perfect Square Problems. By following this article, you will better understand the concept and solving process of perfect square factorization.

(i) a² + 2ab + b² = (a + b)² = (a + b) (a + b)
(ii) a² – 2ab + b² = (a – b)² = (a – b) (a – b)

Factorization of Perfect Square Solved Examples

1. Factorize the perfect square completely
(i) 16a² + 25b² + 40ab

Solution:
Given expression is 16a² + 25b² + 40ab
The given expression 16a² + 25b² + 40ab is in the form a² + 2ab + b².
So find the factors of given expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = 4a, b = 5b
Apply the formula and substitute the a and b values.
16a² + 25b² + 40ab
(4a)² + 2 (4a) (5b) + (5b)²
(4a + 5b)²
(4a + 5b) (4a + 5b)

Factors of the 16a² + 25b² + 40ab are (4a + 5b) (4a + 5b)

(ii) 9x² – 42xy + 49y²

Solution:
Given expression is 9x² – 42xy + 49y²
The given expression 9x² – 42xy + 49y² is in the form a² – 2ab + b².
So find the factors of given expression using a² – 2ab + b² = (a – b)² = (a – b) (a – b) where a = 3x, b = 7y
Apply the formula and substitute the a and b values.
9x² – 42xy +49y²
(9x)² – 2 (9x) (7y) + (7y)²
(9x – 7y)²
(9x – 7y) (9x – 7y)

Factors of the 9x² – 42xy + 49y² are (9x – 7y) (9x – 7y)

(iii) 25m² + 80m + 64

Solution:
Given expression is 25m² + 80m + 64
The given expression 25m² + 80m + 64 is in the form a² + 2ab + b².
So find the factors of given expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = 5m, b = 8
Apply the formula and substitute the a and b values.
25m² + 80m + 64
(5m)² + 2 (5m) (8) + (8)²
(5m + 8)²
(5m + 8) (5m + 8)

Factors of the 25m² + 80m + 64 are (5m + 8) (5m + 8)

(iv) a² + 6a + 8

Solution:
Given expression is a² + 6a + 8.
The Given expression is a² + 6a + 8 is not a perfect square.
Add and subtract 1 to make the given expression a² + 6a + 8 is not a perfect square.
a² + 6a + 8 + 1 – 1
a² + 6a + 9 – 1
The above expression a² + 6a + 9 is in the form a² + 2ab + b².
So find the factors of the expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = a, b = 3
Apply the formula and substitute the a and b values.
a² + 6a + 9
(a)² + 2 (a) (3) + (3)²
(a + 3)²
(a + 3)² – 1
(a + 3)² – (1)²
(a + 3 + 1) (a + 3 – 1)
(a + 4) (a + 2)

Factors of the a² + 6a + 8 are (a + 4) (a + 2)

2. Factor using the identity

(i) 4x⁴ + 1

Solution:
Given expression is 4x⁴ + 1.
The Given expression is 4x⁴ + 1 is not a perfect square.
Add and subtract 4x² to make the given expression 4x⁴ + 1 is not a perfect square.
4x⁴ + 1 + 4x² – 4x²
4x⁴ + 4x² + 1 – 4x²
The above expression 4x⁴ + 4x² + 1 is in the form a² + 2ab + b².
So find the factors of the expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = 2x², b = 1
Apply the formula and substitute the a and b values.
4x⁴ + 4x² + 1
(2x²)² + 2 (2x²) (1) + (1)²
(2x² + 1)²
(2x² + 1)² – 4x²
(2x² + 1)² – (2x)²
(2x² + 1 + 2x) (2x² + 1 – 2x)
(2x² + 2x + 1) (2x² – 2x + 1)

Factors of the 4×4 + 1 are (2x² + 2x + 1) (2x² – 2x + 1)

(ii) (a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)²

Solution:
Given expression is (a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)²
The given expression (a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)² is in the form a² + 2ab + b².
So find the factors of given expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = a + 2b, b = 3b – a
Apply the formula and substitute the a and b values.
(a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)²
(a + 2b)² + 2 (a + 2b) (3b – a) + (3b – a)²
(a + 2b + 3b – a)²
(5b)²
25b²

Factors of the (a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)² are 25b²

Looking for help on the concept Train Passes through a Bridge then you can find all of it here. Refer to Solved Examples on How to find the Speed Time and Distance when a Train Crosses a Bridge. We have provided solutions for all the Problems covered on the concept of Train Passing through a Bridge. or Tunnel or any Stationary Object.

In this case, the Length of the Train and Length of the Bridge are added to get the Distance. Practice the Train Passes through a Bridge Problems with Answers and get the elaborate explanation provided to understand the concept better.

How to Solve Train Passing through a Bridge or Tunnel Problems?

Let us consider the Length of the Train = x meters

Length of the Stationary Object = y meters

The Speed of the Train is z km/hr

Time Taken by the Train to Cross the Bridge = (Length of Train + Length of Bridge)/Speed of Train

= (x+y)m /z km/hr

To change between km/hr to m/sec multiply with 5/18

Solved Examples on Train Crossing a Stationary Object having Some Length

1. A train 250 m long crosses a bridge which is 100 m long in 50 seconds. What is the speed of the train?

Solution:

Length of the Train = 250 m

Length of the Bridge = 100 m

Speed of the Train = (Length of the Train+Length of Stationary Object)/Time

= (250 m+100 m)/50

= 350 m/50 sec

= 7 m/sec

2. A train 260 m long is running at a speed of 40 km/hr. What time will it take to cross an 80 m long tunnel?

Solution:

Length of the Train = 260 m

Speed of the Train = 40 km/hr

= 40 *5/18 m/sec

= 11.11 m/sec

Length of the Tunnel = 80 m

Time = Distance/Speed

= 260 m/11.11 m/sec

= 23.40 sec

Train takes 23.40 sec to travel the tunnel.

3. Find the time taken by a 180 m long train passes through a bridge which is 75 m long, running at a speed of 54 km/hr?

Solution:

Length of the Train = 180 m

Length of the Bridge = 75 m

Speed of the Train = 54 km/hr

Speed of train in m/sec = 54*5/18

= 15

Time = Distance/Speed

= 225 m/15 m/sec

= 9 sec

Therefore, train takes 9 sec to cross the bridge.

4. A 120 m long train is running at a speed of 45 km/hr. If it takes 20 seconds to cross a platform, find the length of the platform?

Solution:

Speed of the Train = 45 km/hr

Speed of Train in m/sec = 45*5/18

= 12.5 m/sec

Time taken to cross the platform = 20 sec

Length of the Train = 120 m

Consider the Length of Platform = x

Speed = (Length of Train +Length of Platform)/Time

12.5 m/sec = (120+x)/20

12.5*20 = 120+x

250 = 120+x

x = 250 – 120

= 130 m

Therefore, Length of Platform is 130 m.

Practice the Questions available below when Train Passes through a Pole and get full-fledged knowledge about the concept. Know How to Calculate Speed Time and Distance when a train crosses a stationary object in the coming modules. Train Passes a Pole, Man or Tree Problems with Solutions along with step by step explanation makes it easy for you to understand the concept. Refer to them and learn how to solve the Train Passing through a Pole Problems easily.

How to Solve Train Passes through a Pole Problems?

When a Train Passes through a Pole or Tree the formulas to find the Time Speed and Distance are given below

Suppose the Length of the Train = x mts

Speed of the Train = y km/hr

Time taken by the Train to Cross a Pole or Stationary Object = Length of the Train/Speed of the Train

= x meters/y km/hr

To Change km/hr to m/sec simply multiply with 5/18.

Solved Examples to Calculate When Train Passes through a Pole

1. A train 200 m long is running at a uniform speed of 75 km/hr. How much time will it take to cross a pole?

Solution:

Length of the Train = 200 m

Speed of the Train = 75 km/hr

To convert it to m/sec multiply with 5/18

= 75*5/18

= 20.83 m/sec

Time taken by the train to cross the pole = Length of the Train/Speed of the Train

= 200 m/20.83 m/sec

= 9.6 sec

Therefore, Train takes 9.6 sec to cross the pole.

2. Find the time taken by a train 350 m long, running at a speed of 60 km/hr in crossing the pole?

Solution:

Length of the Train = 350 m

Speed of the Train = 60 km/hr

Time taken by the Train to cross the pole = Length of the Train/Speed of the Train

= 350m/60 km/hr

To change from km/hr to m/sec multiply with 5/18 i.e. 60*5/18 = 16.66 m/sec

= 350 m/16.66m/sec

= 21.008 sec

Thus, the train takes 21.008 sec to cross the pole.

3. A train is running at a speed of 54 km/hr. It crosses a tower in 7 seconds. Find the Length of the train?

Solution:

Speed of the Train = 54 km/hr

Time taken by train to cross tower = 7 seconds

Length of the Train = Speed * Time

= 54*(5/18)*7

= 105 m

Therefore, the Length of the Train is 105 m

4. A train is running at a speed of 140 km/hr. if it crosses a pole in just 8 seconds, what is the length of the train?

Solution:

Speed of the Train = 140 km/hr

Time taken to cross the pole = 8 seconds

Length of the Train = Speed * Time

= 140 km/hr*8

= 140*(5/18)*8

= 311.11 m

Therefore, the length of the train is 311.11 m

Are you looking for help on the concept Train Passes a Moving Object in the Opposite Direction? Then you have reached the right place. Learn the Formulas for Speed Time and Distance in the case of a train crossing a moving body in the opposite direction. Get Solved Examples for finding the Train Crossing a Moving Object in Opposite Direction long with detailed solutions. Practice the Problems available and get a good hold of the concept.

How to calculate Time Speed and Distance for Train Crossing a Moving Object in Opposite Direction?

Let us assume the length of the train = l m

Speed of the train =  x km/hr

Speed of the object = y km/hr

Relative Speed = (x+y) km/hr

Time taken by train to cross the moving object = Distance/Speed

= l m/(x+y) km/hr

Simply rearrange the formula to obtain the other measures if few are known.

Solved Problems on Train Passes a Moving Object in the Opposite Direction

1. A train 200 m long is running at a speed of 50 km/hr. In what time will it pass a man who is running at the speed of 4 km/hr in the opposite direction in which the train is moving?

Solution:

Length of the Train = 200 m

Speed of the Train = 50 km/hr

Speed of Man = 4 km/hr

Relative Speed = (50+4) Km/hr

= 54 km/hr

Time taken by train to cross a man = Distance/Relative Speed

= 200 m/54 Km/hr

= 200 m/ (54*5/18) m/sec

= 200 m/15 m/sec

= 13.33 sec

Therefore, the train takes 13.33 sec to cross the man.

2. Two trains 125 meters and 170 meters long are running in the opposite direction with speeds of 60 km/hr and 45 km/hr. In how much time they will cross each other?

Solution:

Total Distance Covered = 125+170

= 295 m

Speed of first train = 60 kmph

Speed of second train = 45 kmph

Relative Speed = 60+45

= 105 kmph

Relative Speed in m/sec = 105*5/18

= 29.1 m/sec

Time taken by trains to cross each other = Distance/Speed

=295m/29.1 m/sec

= 10.1 sec

Therefore, it takes 10.1 sec for both the trains to cross each other.

3. Two trains running in opposite directions cross a man standing on the platform in 30 seconds and 21 seconds respectively and they cross each other in 25 seconds. The ratio of their speeds is?

Solution:

Let us consider the speed of two trains be x m/sec and y m/sec

Length of the first train = 30x m(since distance = Speed*Time)

Length of the second train = 21y m

From the given data

Time taken by both trains to cross each other = 25 sec

(30x+21y)/(x+y) = 25

30x+21y = 25x+25y

30x-25x = 25y-21y

5x=4y

x/y = 4/5

Therefore, the Ratio of Speeds is 4:5

Gain complete knowledge on the concept Train Passes a Moving Object in the Same Direction. Learn related formulas like Speed, Time and Distance when a train passes through a moving object in the Same Direction. Get the Step by Step Procedure along with a detailed explanation for the entire concept. Solve Problems on Train Passing a Moving Object in the Same Direction and understand the concept behind them easily.

How to find Time Speed and Distance when a Train Passes a Moving Object in the Same Direction?

Follow the guidelines for calculating the Time Speed and Distance when a Train Passes a Moving Body in the Same Direction. They are as such

Let us consider the length of the train as l mt and the speed of the train is x km/hr

Speed of the Object = y km/hr

Relative Speed = (x-y) km/hr

Time Taken by the train to cross a moving object in the same direction is = Distance/ Relative Speed

= l m/(x-y) km/hr

You can rearrange the equation and find whichever measure you need as a part of your work.

Solved Problems on Train Passes through a Moving Object in the Same Direction

1. A train 150 m long is running at a speed of 50 km/hr. At what time will it pass a man who is running at the speed of 5 km/hr in the same direction in which the train is moving?

Solution:

Length of the Train = 150 m

Speed of the Train = 50 Km/hr

Speed of the Man = 5 km/hr

Relative Speed = Speed of Train – Speed of Man

= 50 – 5

= 45 km/hr

= 45 *5/18

= 12.5 m/sec

Time Taken by Train to Cross the Man = Distance/Speed

= 150 m/12.5 m/sec

= 12 sec

Therefore, Train takes 12 sec to cross the man.

2. Two trains 110 meters and 140 meters long are running in the same direction with speeds of 70 km/hr and 55 km/hr. In how much time will the first train cross the second?

Solution:

Distance Covered = 110+140

= 250 meters

Speed of first train = 70 km/hr

Speed of second train = 55 km/hr

Relative Speed = (70 – 55)

= 15 km/hr

Relative Speed in m/sec = 15*5/18

= 4.1 m/sec

Time taken by first train to cross second = Distance/Speed

= 250 m/4.1 m/sec

= 60.9 sec

Therefore, the first train takes 60.9 sec to cross the second train.

3. A train running at 60 kmph takes 30 seconds to pass a platform. Next, it takes 10 seconds to pass a man walking at 5 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform?

Solution:

Let us consider the length of the train and length of the platform as x and y

Distance traveled by train while crossing the platform is x+y

Time taken to cross the platform is 30 sec

Speed of the train = 60 kmph

= 60 *5/18

= 16.66 m/sec

Time taken by train to cross the platform is

Time = Distance/Speed

30 =(x+y)/16.66  …….(1)

Time taken by train to cross the platform next = 10 sec

Speed of the man = 5 kmph

Relative Speed = 60 kmph – 5 kmph

= 55 kmph

= 55 *5/18

= 15.2 m/sec

Time taken by train to cross the man is

Time = Distance/Speed

10 sec = x/(15.2 m/sec)

x =  152.7 m

By applying the value of x in equation 1 we have

30 =(152.7+y)/16.66

499.8 = 152.7+y

y = 347.1 m

Hence the length of the train and platform are 152.7 m and 347.1 m respectively.

Get full-fledged knowledge when two objects move in opposite direction by going through the complete article. Learn Formula, Solved Examples on Two Bodies Moving in Opposite Direction towards each other. Get Step by Step Solutions for all the Problems on finding Relative Speed, Time and Distance. If two bodies move in opposite direction relative speed is obtained by adding the speeds of both the bodies.

How to Calculate Time Speed and Distance for Two Objects Moving in Opposite Direction?

Consider Two Objects Moving in Opposite Direction with different speeds

Speed of 1st Object = x km/hr

Speed of 2nd Object = y km/hr

Relative Speed = (x+y) km/hr

The speed of One Object with respect to another is called relative speed.

Distance between them = d km

Time after two objects meet = d km/(x+y) km/hr

Distance covered in t hours = relative speed * time

= (x+y) km/hr * t

Solved Examples on Two Object Moving in Opposite Direction

1. Two Athletes are running from the same place at the speed of 10 km/hr and 5 km/hr. find the distance between them after 20 minutes if they move in the opposite direction?

Solution:

Speed of 1st Athlet = 10 km/hr

Speed of 2nd Athlet = 5 km/hr

Relative Speed = (10+5) km/hr

= 15 km/hr

Time = 20 min = 20/60 = 1/3 hr

Distance = Relative Speed * Time

= 15 km/hr * 1/3 hr

= 5 km

Distance between Two Athlets is 5 Km.

2. Two cars travel from the same location at the speed of 10 km/hr and 5 km/hr respectively. Calculate the distance between the cars after 30 minutes given that both cars are traveling in the opposite direction?

Solution:

Speed of 1st Car = 10 km/hr

Speed of 2nd Car = 5 km/hr

Time = 30 minutes = 1/2 hr

Relative Speed of Cars = (10+5) km/hr

= 15 km/hr

Distance between Cars = Relative Speed * Time

= 15 km/hr * 1/2 hr

= 7.5 km

Both cars have a distance of 7.5 km

When Two Objects move in the same direction then their relative speed is calculated by the difference of their speeds. Get to know about the Time and Distance Formulas when two objects move in the same direction. Get to all about the concept objects that move in the same direction by going through the entire article. Learn, how to calculate the Speed when two objects travel in the same direction by checking the solved examples provided.

How to Calculate Speed Time and Distance when Two Objects Move in the Same Direction?

Let us assume two bodies or objects are moving in the same direction having different speeds.

Suppose the Speed of 1st Object is x km/hr

Speed of 2nd Object is y km/hr

Thus, the Relative Speed = (x – y) km/hr[if x>y]

Time taken by two objects to meet = Distance Traveled/Relative Speed

= d km/(x-y) km/hr

We know that Relative Speed is the Speed of the object with respect to one another

Consider the taken after both the bodies meet = t hrs

Distance covered in t hrs = Time * Relative Speed

= (x-y)km/hr *t hrs

For better understanding refer to the solved problems explaining how to calculate when two objects move in the same direction.

Solved Problems on Two Objects Moving in the Same Direction

1. Two athletes are running from the same place at the speed of 8 km/hr and 6 km/hr. find the distance between them after 20 minutes if they move in the same direction?

Solution:

Speed of Athlet A = 8 km/hr

Speed of Athlet B = 6 km/hr

Time = 20 mins = 1/3 hr

Relative Speed = (8-6) km/hr = 2 km/hr

Distance traveled by them = Relative Speed of Athlets/Time Taken

= 2 km/hr/(1/3) hr

= 6 km

Distance between after they travel for 20 mins is 6 Km.

2. Two vehicles are traveling from the same location at the speed of 10 km/hr and 7 km/hr respectively. Calculate the distance between the vehicles after 15 minutes given that both vehicles are traveling in the same direction?

Solution:

Speed of first Vehicle = 10 km/hr

Speed of second Vehicle = 7 km/hr

Relative Speed = (10-7) km/hr

= 3 km/hr

Time = 15 minutes = 15/60 = 1/4 hr

Distance between both the vehicles = Relative Speed * Time

= 3km/hr *1/4 hr

= 3/4 km

= 0.75 km

Therefore, the distance between both the vehicles is 0.75 km

Learn completely on how to convert from one unit of speed to others from here. Practice the Questions in Conversion of Units of Speed here and get a good hold of the concept. Convert from km/hr to m/sec and m/sec to km/hr easily by going through the further sections. To know more about Speed Time and Distance you can always look up to us. Check out Solved Examples on Speed Conversions so that you can solve related problems on your own. Step by Step solutions provided makes it easy for you to understand the Units of Speed Conversion in an effective manner.

How to Convert Km/hr to m/sec?

To convert Km/hr to m/sec follow the below listed guidelines. They are as such

We know 1 km = 1000m and 1 hour = 60 minutes and in turn 1 minute = 60 sec

1 hour = 60*60 = 3600 sec

Dividing km/hr we have = 1000m/3600sec

On simplifying we have km/hr = 5/18 m/sec

Therefore to convert from km/hr we simply multiply with 5/18.

Solved Problems on Converting Km/hr to m/sec

1. Convert 54 km/hr to m/sec?

Solution:

To convert, 54 km/hr to m/sec simply multiply with 5/18

= 54*5/18

= 15 m/sec

Therefore 54km/hr converted to m/sec is 15m/sec.

2. The speed of the bike is 108 km/hr, what is its speed in m/sec?

Solution:

Given Speed of Bike = 108 km/hr

To obtain the Speed in m/sec simply multiply with 5/18

= 108*5/18

= 30 m/sec

Therefore, the Speed of the Bike is 30 m/sec.

3. A car covers a distance of 120 km in the first three hours, 50 km in the next 1 hour, and 25 km in the next 1/2 hour. Convert the speed into m/sec?

Solution:

Total Distance traveled by the Car = (120+50+25) = 195 km

Total Time Taken = (3+1+1/2) = 4 1/2 hrs

Speed of the Car = 195km/4.5 hr = 43.3 km/hr

Speed of the Car in m/sec = 43.3*18/5

= 155.88 m/sec

How to Convert m/sec to km/hr?

Follow the below listed guidelines inorder to change between m/sec to km/hr

We know 1km = 1000m

thus, 1m = 1/1000 km

1 hour = 3600 sec

thus, 1 sec = 1/3600 hr

applying the same we have 1 m/1 sec = (1/1000km)/(1/3600 hr)

= 3600km/1000hr

= 18/5 km/hr

Therefore, to convert from m/sec to km/hr simply multiply with 18/5

Solved Problems on Converting m/sec to km/hr

1. Convert 60 m/sec to km/hr?

Solution:

Given 60 m/sec

to change to km/hr multiply with 18/5

= 60*18/5

= 216 km/hr

2. The speed of a cyclist is 15 m/sec. find the speed in km/hr?

Solution:

Speed of a Cyclist = 15 m/sec

To convert to km/hr multiply with 18/5

= 15*18/5

= 54 km/hr