Common Core Math

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).

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)