# Factoring Trinomials (Quadratics) : Lucid Explanation of the Method With Examples

09:15 2009

We give step by step explanation of the method of factoring Quadratic Expressions. We apply the step by step method to solve a number of examples. The examples are so chosen that all the models are covered. Lucid Explanation of the method and its application to solve a number of problems, help the reader to feel confident to solve similar problems on his own.

Consider the product of the two linear expressions (y+a) and (y+b).

(y+a)(y+b) = y(y+b) + a(y+b) = y^2 + by + ay + ab = y^2 + y(a+b) + ab

We can write it as

y^2 + y(a+b) + ab = (y+a)(y+b) .......(i)

Similarly, Consider the product of the two linear expressions (ay+b) and (cy+d).

(ay+b)(cy+d) = ay(cy+d) + b(cy+d) = acy^2 + ady + bcy + bd = acy^2 + y(ad+bc) + bd

We can write it as

acy^2 + y(ad+bc) + bd = (ay+b)(cy+d) .......(ii)

Equation (i) is Simple Quadratic Polynomial expressed as Product of Two linear Factors

and Equation (ii) is General Quadratic Polynomial expressed as Product of Two linear Factors

Observing the two Formulas, leads us to the method of Factorization of Quadratic Expressions.

In Equation (i),

the product of coefficient of y^2 and the constant term = ab

and the coefficient of y = a+b = sum of the factors of ab

Similarly, In Equation (ii),

the product of coefficient of y^2 and the constant term = (ac)(bd) = (ad)(bc)

and the coefficient of y = (ad+bc) = sum of the factors of acbd

So, if we can resolve the product of y^2 and the constant term into product of two factors in such a way that their sum is equal to the coefficient of y, then we can factorize the quadratic expression.

We discuss the steps involved in the method and apply it to solve a number of problems.

Method of Factoring Trinomials (Quadratics) :

Step 1 :

Multiply the coefficient of y^2 by the constant term.

Step 2 :

Resolve this product into two factors such that their sum is the coefficient of y

Step 3 :

Rewrite the y term as the sum of two terms with these factors as coefficients.

Step 4 :

Then take the common factor in the first two terms and the last two terms.

Step 5 :

Then take the common factor from the two terms thus formed.

What you get in step 5 is the product of the required two factors.

The method will be clear by the following Solved Examples.

The examples are so chosen that all the models are covered.

Example 1 :

Factorize 9y^2 + 26y + 16

Solution :

Let P = 9y^2 + 26y + 16

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = 9 x 16 = 144

Step 2:

We have to express 144 as two factors whose sum = coefficient of x = 26;

144 = 2 x 72 = 2 x 2 x 36 = 2 x 2 x 2 x 18 = 8 x 18; (8 + 18 = 26)

Step 3:

P = 9y^2 + 26y + 16 = 9y^2 + 8y + 18y + 16

Step 4:

P = y(9y + 8) + 2(9y + 8)

Step 5:

P = (9y + 8)(y + 2)

Thus, 9y^2 + 26y + 16 = (9y + 8)(y + 2) Ans.

Example 2 :

Factorize y^2 + 7y - 78

Solution :

Let P = y^2 + 7y - 78

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = 1 x -78 = -78

Step 2:

We have to express -78 as two factors whose sum = coefficient of y = 7 ;

-78 = -2 x 39 = -2 x 3 x 13 = -6 x 13; (-6 + 13 = 7)

Step 3:

P = y^2 + 7y - 78 = y^2 - 6y + 13y - 78

Step 4:

P = y(y - 6) + 13(y - 6)

Step 5:

P = (y - 6)(y + 13)

Thus, y^2 + 7y - 78 = (y - 6)(y + 13) Ans.

Example 3 :

Factorize 4y^2 - 5y + 1

Solution :

Let P = 4y^2 - 5y + 1

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = 4 x 1 = 4

Step 2:

We have to express 4 as two factors whose sum = coefficient of y = -5 ;

4 = 4 x 1 = -4 x -1; [(-4) + (-1) = -5]

Step 3:

P = 4y^2 - 5y + 1 = 4y^2 - 4y - y + 1

Step 4:

P = 4y(y - 1) - 1(y - 1)

Step 5:

P = (y - 1)(4y - 1)

Thus, 4y^2 - 5y + 1 = (y - 1)(4y - 1) Ans.

Example 4 :

Factorize 3y^2 - 17y - 20

Solution :

Let P = 3y^2 - 17y - 20

Now, follow the five steps listed above.

Step 1:

Coefficient of y^2 x constant term = 3 x -20 = -60

Step 2:

We have to express -60 as two factors whose sum = coefficient of x = -17 ;

-60 = -20 x 3; (-20 + 3 = -17)

Step 3:

P = 3y^2 - 17y - 20 = 3y^2 - 20y + 3y - 20

Step 4:

P = y(3y - 20) + 1(3y - 20)

Step 5:

P = (3y - 20)(y + 1)

Thus, 3y^2 - 17y - 20 = (3y - 20)(y + 1) Ans.

Example 5 :

Factorize 2 - 5y - 18y^2

Solution :

Let P = 2 - 5y - 18y^2 = -18y^2 - 5y + 2

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = -18 x 2 = -36

Step 2:

We have to express -36 as two factors whose sum = coefficient of y = -5 ;

-36 = -2 x 18 = -2 x 2 x 9 = 4 x -9; [4 + (-9) = -5]

Step 3:

P = -18y^2 - 5y + 2 = -18y^2 + 4y - 9y + 2

Step 4:

P = 2y(-9y + 2) + 1(-9y + 2)

Step 5:

P = (-9y + 2)(2y + 1)

Thus, 2 - 5y - 18y^2 = (-9y + 2)(2y + 1) Ans.

Example 6 :

Factorize (y^2 + y)^2 -18(y^2 + y) + 72

Solution :

Let P = (y^2 + y)^2 -18(y^2 + y) + 72

Put (y^2 + y) = t; Then P = t^2 -18t + 72

Now, follow the five steps listed above.

Step 1:

(Coefficient of t^2) x (constant term) = 1 x 72 = 72

Step 2:

We have to express 72 as two factors whose sum = coefficient of t = -18 ;

72 = 12 x 6 = -12 x -6; [(-12) + (-6) = -18]

Step 3:

P = t^2 -18t + 72 = t^2 - 12t - 6t + 72

Step 4:

P = t(t - 12) - 6(t - 12)

Step 5:

P = (t - 12)(t - 6)

But t = (y^2 + y);

So, P = (t - 12)(t - 6) = (y^2 + y - 12)(y^2 + y - 6)

In each of these two brackets, there is a Quadratic Polynomial which can be factorised using the five steps above.

y^2 + y - 12 = y^2 + 4y - 3y - 12 = y(y + 4) - 3(y + 4) = (y + 4)(y - 3)

y^2 + y - 6 = y^2 + 3y - 2y - 6 = y(y + 3) - 2(y + 3) = (y + 3)(y - 2)

See how these two Quadratic Polynomials are factorised with the knowledge of the 5 steps.

You might have mastered the 5 steps of factorisation by this time, to write directly like this.

Thus,

P = (y^2 + y)^2 -18(y^2 + y) + 72

= (y^2 + y - 12)(y^2 + y - 6)

= (y + 4)(y - 3)(y + 3)(y - 2) Ans.

For more, on Factoring Quadratics, go to

http://www.math-help-ace.com/Factoring-Trinomials.html

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