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Practice tracing the numbers from 21 to 30

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Practice tracing the numbers from 41 to 50

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Missing Number Worksheets

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Worksheet on Before, After, and Between Numbers up to 20
Worksheet on Before, After, and Between Numbers up to 50

Factorization of Quadratic Trinomials | How to Find the Factors of a Quadratic Trinomial?

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).

Factorize the Trinomial ax^2 + bx + c | How to Factor a Trinomial in the Form ax^2 + bx + c?

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).

Factorize the Trinomial x^2 + px + q | How to Find Factorization of Trinomial of the Form x^2 + px + q?

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).

Factorization of Perfect Square Trinomials | How to Factor Perfect Square Trinomials?

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)