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# Solving by Factoring

In algebra, factoring is a process of finding the factors. This is a process to split complex expressions into a multiplication of simpler expressions.

For Example: quadratic polynomial, 12($x^2-1$) can be splitted as 2 $\times$ 2 $\times$ 3$(x-1)(x+1)$. Factoring and expanding are two different concepts which are used in mathematics.

Expanding process is opposite of factoring. This process is mainly used in finding the LCF, GCF and in polynomials factorization. There are various algebraic formula to factoring the polynomials. Factoring polynomial expressions is quite different from factoring numbers. Factoring is the opposite process of multiplying polynomials.

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## How to Solve Polynomial by Factoring Method

Solving polynomial by factorization is the opposite process of multiplying polynomials. When we factor a polynomial, we get simpler polynomials that can be multiplied together to give us the given polynomial. Quadratic formula, synthetic division and long division methods help to find the factors of polynomials.

Follow the below steps to find the factors of a polynomial:
Step 1: Write the expression in the correct form. And find the common terms between given terms.

For example, $2x^3+12x^2-54x = 0$

Step 2: Take out common terms and write rest inside the bracket.

$2x^3+12x^2-54x= 0$

or $2x(x^2+6x-27)= 0$

Step 3: Use a factoring strategies to factor further (If possible).

$2x(x^2+6x-27)= 0$

or 2x (x - 3)(x + 9) = 0

Step 4: Use zero product principle.

2x = 0, (x - 3) = 0, (x + 9) = 0

x = 0, x = 3 and x = -9 is the final answer.

While factoring a polynomial usually we break it down into simplest form. For example, Factor 10y$^2$ - 20y - 30, here 10 is common in all the three terms.

10y$^2$ - 20y - 30 = 10(y$^2$ - 2y - 3)

10(y$^2$ - 2y - 3) is the answer.

## Factor of Quadratic Polynomial

A quadratic polynomial is a polynomial of degree two. It is also called as second degree polynomial due to the highest exponent of variable is two. It can be represented in the form, g(y) = ay$^2$ + by + c, where a $\neq$ 0. Quadratic polynomial has two roots and it can be calculated by using formula, y = $\frac{-b \pm \sqrt{D}}{2a}$, where D = $b^2$ - 4ac is the discriminant of the polynomial.
To find the factors of a quadratic polynomial, ay$^2$ + by + c, have to follow below steps:
Step 1: Compare coefficients of given polynomial with ay$^2$ + by + c.

Step 2: Find the factors of ac.

Step 3: Choose two numbers that will not only multiply to equal the product of a and c, but also add/subtract equals to b.
For Example: Factorize $2y^2+17y-9$

Compare given polynomial with ay$^2$ + by + c, we get

a = 2, b = 17 and c = -9

ac = -18

Choose 17 and -1 as factors of -18 because 18 - 1 = 17 = b and 18 * -1 = -18 = ac

Split 17y as 18y - y

$2y^2+17y-9$ = $2y^2+18y-y-9$

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

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

Therefore factors of $2y^2+17y-9$ are (2y - 1) and (y + 9).

## Factoring of Trinomials

In algebra, a trinomial is a polynomial consisting of three monomials. For example 2x + y + 6, ax + st + uv, ax$^2$ + ut + mn$^2$ all are trinomials. Trinomials can be factorized as same as we factorized other polynomials.

Synthetic division and long division methods are really helpful to factorize trinomial if we know any one root of the given expression.

Example: Factor the trinomial $x^2$ + 2x - 63

Solution: $x^2$ + 2x - 63 = $x^2$ + 9x - 7x - 63

= x (x + 9) - 7(x + 9)

= (x - 7)(x + 9)

factors of given trinomial are (x - 7) and (x + 9).

## Examples of Factoring

Below are some solved examples which help to understand better how to find factors of polynomials.
Example 1: Find the factors of $x^3 + 2x^2-x-2$. Given that x - 1 is one of the factors.

Solution: Since x - 1 is one of the factors of $x^3 + 2x^2-x-2$. Let us reduce this polynomial with the help of long division method.

$x^3 + 2x^2-x-2$ = (x - 1)($x^2 + 3x+2$)   ..........(1)

Again we can split the trinomial ($x^2 + 3x+2$) into simpler factors.

$x^2 + 3x+2$ = $x^2 + 2x + x +2$ = x(x + 2) + (x + 2)

(x + 1)(x + 2)

Therefore (x - 1), (x + 1) and (x + 2) are factors of $x^3 + 2x^2-x-2$.

Example 2:  Factor the trinomial, 3x$^2$ + x - 10

Solution: 3x$^2$ + x - 10

Compare given polynomial with ay$^2$ + by + c, we get

a = 3, b = 1 and c = -10

ac = -30

Choose 6 and -5 as factors of -30 because 6 - 5 = 1 = b and 6 * -5 = -30 = ac

Split x as 6x - 5x

3x$^2$ + x - 10 = 3x$^2$ + 6x - 5x - 10

= 3x(x + 2) - 5(x + 2)

= (3x - 5)(x + 2)

(3x - 5) and (x + 2) are the factors of 3x$^2$ + x - 10.

Example 3: Factor completely 40xy + 100 + 10$x^3$

Solution: Rearrange 40xy + 100 + 10$x^3$ as 10$x^3$ + 40xy + 100x

Find common terms between 3 terms

10$x^3$ + 40xy + 100x = 10x(x$^2$ + 4y + 10)

Example 4: Find  the factors of $y^3-5y^2+2y+8$

Solution: Factors of constant term i.e. 8 = 1, 2, 4, 8

Let f(y) = $y^3-5y^2+2y+8$

To find first factor of given polynomial substitute all the factors of 8 in f(y).

f(1) $\neq$ 0

f(2) = 0 ( y = 2 is one of the factors)

Now let us apply synthetic division to find rest of the factors.

We are left with quadratic polynomial, y$^2$ - 3y - 4

y$^2$ - 3y - 4 = y$^2$ - 4y + y - 4 = y(y - 4) + (y - 4) = (y + 1)(y - 4)

$\therefore$ Factors of given polynomial are (y - 2)(y + 1)(y - 4) .
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