"**Polynomial**" word is composed of two individual terms - **poly **means **many **and **nomial **means **terms**. Thus, a polynomial does usually have more than one terms. Polynomials are defined as expressions which are formed by addition or subtraction of variables and constants, known as **monomials. **Polynomial of an order **n** can be represented in the following general form -

P(x) = $a_{n}x^{n}+a_{n-1}x^{n-1}+...+ a_{0}$

where, $a_{n}$ is not equal to zero.Any polynomial of the form F(a) can also be written as

P(x) = Q(x) * D (x) + R

[Dividend = Quotient * Divisor + Remainder]

If the polynomial F(x) is fully divisible by Q(x), then the remainder will be zero. Thus, F(x) = Q(x) * D(x). Thus, the polynomial F(x) is a product of two other polynomials Q(x) and D(x). There are various different methods of factoring polynomials according to their types and forms. Let us learn in brief what these technique are and in what manner these techniques are applied to which type of polynomial.

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Steps for factoring polynomials:

Step 1: Find common factors or GCF.

Step 2: Use suitable method to solve polynomial. Like

For four terms: Factor by grouping.

Quadratic polynomial can be factored using the following three methods:

**Use the below widget to factor polynomia****l expression.**

A cubic polynomial is a polynomial of degree 3. The cubic polynomial, ax^{3} + bx^{2} + cx + d, where a, b, c and d are real coefficients and a $\neq$ 0. Let see with the help of example how to find the factors of cubic polynomials.

Let us find the solution set of x^{3 }- 2x^{2} - 2x - 3.

**Step 1:**

Given cubic polynomial is x^{3 }- 2x^{2} - 2x - 3

=> Possible roots of the polynomial are $\pm$ 1, $\pm$ 3.

Step 2:

To find root, plug each value into polynomial

Put x = -1

=> x^{3 }- 2x^{2} - 2x - 3 = (-1)^{3 }- 2(-1)^{2} - 2(-1) - 3 = -1 - 2 + 2 - 3 $\neq$ 0

=> x = -1 is not a root.

Put x = 1

=> x^{3 }- 2x^{2} - 2x - 3 = 1^{3 }- 2 * 1^{2} - 2 * 1 - 3 = 1 - 2 - 2 - 3 $\neq$ 0

=> x = 1 is not a root.

Put x = 3

=> x^{3 }- 2x^{2} - 2x - 3 = 3^{3 }- 2 * 3^{2} - 2 * 3 - 3 = 27 - 18 - 6 - 3 = 27 - 27 = 0

=> x = 3 is a root.

Step 3:

Find other roots using synthetic division.

=> x^{3 }- 2x^{2} - 2x - 3 = (x - 3)(x^{2} + x + 1) .....................(i)

The factors of x^{2} + x + 1 are (x - $\frac{-1 + \sqrt{3}i}{2}$) and (x - $\frac{-1 - \sqrt{3}i}{2}$ )

Therefore, the factors of x^{3 }- 2x^{2} - 2x - 3 are (x - 3), (x - $\frac{-1 + \sqrt{3}i}{2}$) and (x - $\frac{-1 - \sqrt{3}i}{2}$ ).

The degree of a polynomial is the highest degree of its terms. Polynomials of degree 1 are called linear polynomial, polynomials of degree 2 are called quadratic polynomials and polynomials of degree n are called nth degree polynomials.

Let us find the factors of the polynomial, x^{4} - 5x^{2 }+ 4.

Given x^{4} - 5x^{2 }+ 4

(x^{2})^{2} - 5x^{2 }+ 4

Substitute x^{2} = t

=> (x^{2})^{2} - 5x^{2 }+ 4 = t^{2} - 5t^{ }+ 4

Factor the quadratic

t^{2} - 5t^{ }+ 4 = t^{2} - 4t^{ }- t + 4

= t(t - 4) - (t - 4)

= (t - 1)(t - 4)

To find the answer in x, put t = x^{2}.

=> x^{4} - 5x^{2 }+ 4 = (x^{2} - 1)(x^{2} - 4) = (x^{2} - 1^{2})(x^{2} - 2^{2})

= (x - 1)(x + 1)(x - 2)(x + 2)

Therefore, roots of quartic polynomial are $\pm$ 1, $\pm$ 2.

Roots are the solution to the polynomial equation, which makes the polynomial equal to zero. The number "a" is the root of the polynomial equation if P(a) = 0.

### Factoring 3^{rd} Degree Polynomials

Let us solve the polynomial equation, x^{3 }+ x^{2} - 20x = 0

Given x^{3 }+ x^{2} - 20x = 0

=> x^{3 }+ x^{2} - 20x = x( x^{2 }+ x^{} - 20) = 0 (Take out common term, x)

=> x( x^{2 }+ x - 20) = 0

=> x( x^{2 }+ 5x - 4x - 20) = 0

=> x(x(x + 5) - 4(x + 5)) = 0

=> x(x - 4)(x + 5) = 0

Set each factor equal to zero

=> x = 0, x - 4 = 0 or x + 5 = 0

The values of x are 0, 4, -5.

Therefore, zeros of x^{3 }+ x^{2} - 20x are 0, 4, -5.

Below you could see some problems based on factoring polynomials.

Example 1: Factorize 3x + xy + 3y + y^{2}

Given 3x + xy + 3y + y^{2}There is no factor common to all the terms. So regroup the terms of the expression. In this expression, there is a common factor for the first two terms. Similarly the last two terms have a common factor.

=> 3x + xy + 3y + y^{2 }= x (3 + y) + y ( 3 + y)

=> 3x + xy + 3y + y^{2} = ( 3 + y) ( x + y).

Given below are some of the practice problems on factoring polynomials.

**Problem 1:** Factorize $x^3 + x^2 - 9x - 9$

**Problem 2:** Factorize $x^2 + 8x + 15$

Step 1: Find common factors or GCF.

Step 2: Use suitable method to solve polynomial. Like

For two terms:

Difference of square pattern (a^2 - b^2 = (a - b)(a + b)).

For three terms: ax^2+bx+c

If coefficient of x^2 is 1, then find the factors of c that sum to b.

If coefficient of x^2 is not 1, then find the factors of ac that sum to b.

For four terms: Factor by grouping.

Step 3: Check whether the factors themselves can be factored.

Step 4: Verify the results by multiplying the factors.

**1.** Factoring by grouping

**2.** Completing the square

**3.** Using quadratic formula.

Some of the identities used while factoring polynomials are

- a
^{2 }+ 2ab + b^{2 }= (a + b)^{2} - a
^{2}- 2ab + b^{2}= (a - b)^{2} - a
^{2 }- b^{2 }= (a + b)(a - b) - x
^{2}+ x(a + b) + ab = (x + a)(x + b) - a
^{3}+ b^{3}= (a + b)(a^{2 }- ab + b^{2}) - a
^{3}- b^{3}= (a - b)(a^{2}+ ab + b^{2})

A cubic polynomial is a polynomial of degree 3. The cubic polynomial, ax

Given cubic polynomial is x

=> Possible roots of the polynomial are $\pm$ 1, $\pm$ 3.

Step 2:

To find root, plug each value into polynomial

Put x = -1

=> x

=> x = -1 is not a root.

Put x = 1

=> x

=> x = 1 is not a root.

Put x = 3

=> x

=> x = 3 is a root.

Step 3:

Find other roots using synthetic division.

=> x

The factors of x

Therefore, the factors of x

The degree of a polynomial is the highest degree of its terms. Polynomials of degree 1 are called linear polynomial, polynomials of degree 2 are called quadratic polynomials and polynomials of degree n are called nth degree polynomials.

Let us find the factors of the polynomial, x

Given x

(x

Substitute x

=> (x

Factor the quadratic

t

= t(t - 4) - (t - 4)

= (t - 1)(t - 4)

To find the answer in x, put t = x

=> x

= (x - 1)(x + 1)(x - 2)(x + 2)

Therefore, roots of quartic polynomial are $\pm$ 1, $\pm$ 2.

Roots are the solution to the polynomial equation, which makes the polynomial equal to zero. The number "a" is the root of the polynomial equation if P(a) = 0.

Let us solve the polynomial equation, x

=> x

=> x( x

=> x( x

=> x(x(x + 5) - 4(x + 5)) = 0

=> x(x - 4)(x + 5) = 0

Set each factor equal to zero

=> x = 0, x - 4 = 0 or x + 5 = 0

The values of x are 0, 4, -5.

Therefore, zeros of x

Below you could see some problems based on factoring polynomials.

Example 1: Factorize 3x + xy + 3y + y

=> 3x + xy + 3y + y

=> 3x + xy + 3y + y

Example 2:

Find the factors of polynomial, $x^3 + 2x^2 - x - 2$.

Solution:

Let $F(x)$ = $x^3 + 2x^2 - x - 2$

$\Rightarrow$ Possible roots of the polynomial are $\pm 1,\ \pm 2$.

Step 1:

To find first root, plug each value into polynomial

Put $x$ = $1$

$\Rightarrow$ $F(1)$ = $1^3 + 2 \times 1^2 - 1 - 2 = 1 + 2 - 1 - 2$ = $0$

So, $x - 1$ is a root of $F(x)$

Step 2:

Find other roots using long division

Factoring Polynomials Problems

$\Rightarrow\ x^3 + 2x^2 - x - 2 = (x - 1)(x^2 + 3x + 2)$

Fcatorized the quadratic, $x^2 + 3x + 2$

$\Rightarrow\ x^3 + 2x^2 - x - 2 = x^2 + 2x + x + 2$

= $x(x + 2) + (x + 2)$

= $(x + 1)(x + 2)$

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

Example 3:

Factorize $x^2 - 2$.

Solution:

Given $x^2 - 2$

$x^2 - 2$ = $x^2 - (\sqrt{2})^2$

= $(x - \sqrt{2})(x + \sqrt{2})$

Therefore, factors of $x^2 - 2$ are $(x - \sqrt2)$ and $(x + \sqrt{2})$.

More topics in Factoring Polynomials | |

Solving by Factoring | Rational Roots Theorem |

Difference of Two Squares | Square of a Binomial |

Translations of a Graph | |

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