 Top

Factoring Polynomials Calculator
Top

Factoring Polynomials Calculator in simple terms called as polynomial calculator is an online tool to find the factors of a polynomial. It is a polynomial factoring calculator that makes calculation easy and fun. If any polynomial in the form of Ax2 + Bx + C then it can easily find the factors of that polynomial. Factoring polynomial calculator is ready tool that acts a factoring calculator with steps and answer.

It can take any polynomial, such as monomial or binomial, so it can be used for factoring monomials as well as well factoring binomials.

It can also be referred as Factoring Monomials Calculator or Factoring Binomial Calculator.

Below is given a default polynomial expression, click "Factor". It will find the factors of the given polynomial with steps.

## Steps for Factoring Polynomials

Step 1 :

Observe the values of a, b, and c; a is the coefficient of x2, b is the coefficient of x, and c is the constant.

Step 2 :

Find the discriminant by using the formula (D) = b2 - 4ac.

Step 3 :

If the discriminant is positive(D > 0) , the expression has two real solutions given by. x$_1$ = $\frac{(-b) + \sqrt{D}}{2a}$and x$_2$ =$\frac{(-b) - \sqrt{D}}{2a}$

Step 4 :

Factors are (x - x$_1$)(x - x$_2$)

Step 5 :

Now you can write the final answer.

## Examples on Factoring Polynomials

1. ### Factor x2+3x-4?

Step 1 :

So in the above equation, a(coefficient of x2)=1

And b(coefficient of x)=3

And c(constant)=-4.

Step 2 :

So D= b2 - 4ac. D= 32- 4(1)(-4). D= 9 - (-16). D= 25.

Step 3 :

Now D > 0, the equation has two real solutions given by: x = $\frac{-b \pm \sqrt{D}}{2a}$
x = $\frac{-3 \pm \sqrt{25}}{2}$
x = $\frac{-3 \pm 5}{2}$

x$_1$ = $\frac{-3 + 5}{2}$ = 1
x$_2$ = $\frac{-3 - 5}{2}$ = -4

Step 4 :

So factors are (x-1)(x+4)

(x-1)(x+4)

2. ### Factor x^2+5x=-6 ?

Step 1 :

So D= b2 - 4ac. D=52 - 4(1)(6). D=25 - 24. D= 1.

Step 2 :

Since D > 0, the equation has two real solutions given by. x = $\frac{-b \pm \sqrt{D}}{2a}$
x = $\frac{-5 \pm \sqrt{1}}{2}$
x = $\frac{-5 \pm 1}{2}$

x$_1$ = $\frac{-5 + 1}{2}$ = -2
x$_2$ = $\frac{-5 - 1}{2}$ = -3

Step 3 :

So factors are (x+2)(x+3).

(x+2)(x+3)

3. ### A courier company planned to design a new courier box to pack different industrial goods whose capacity is five times more than its biggest box. Its biggest box is 4 feet long, 3 feet wide, and 2 feet high. The new courier box must be designed by increasing all dimensions by the same measure. Find the increase in each dimension.

Step 1 :

The dimensions of given box = 4 feet  x 3 feet x 2 feet
Volume of box = length * width * height = 4 * 3 * 2 = 24
Since volume of new courier box = 5 (volume of biggest box) = 5(24) = 120

Step 2 :

Let each dimension is increased by same amount, say x. Then new dimensions are as follow:
Length = (x + 2), Width = (x + 3) and Height = (x + 4)

Step 3 :

Now (x + 2)(x + 3)(x + 4) = 120
x$^3$ + 9x$^2$ + 26x + 24 = 120
x$^3$ + 9x$^2$ + 26x - 96 = 0
Let’s apply Hit and Trial method to find the first zero of above cubic equation.
First check result for x = 1, -1, 2, -2.
And we find above equation is true for x = 2. So x = 2 is one of the roots.
If we divide x$^3$ + 9x$^2$ + 26x - 96 by (x - 2) we get quotient as x$^2$  11x + 48.
or x$^3$ + 9x$^2$ + 26x - 96 = (x - 2)(x$^2$ + 11x + 48)

Step 4 :

If we further solve x$^2$ + 11x + 48, we get complex roots i.e x = $\frac{\pm i(\sqrt{71} +11i}{2}$. So let us go ahead with factor (x - 2).
Or x – 2 = 0, x = 2.

Each dimension is increased by 2 feet.

4. ### The sum of the two sides of a rectangular shaped credit card is 7 and the product is 10. Find the dimensions of the credit card.

Step 1 :

Let x be one of the sides of a credit card.

Step 2 :

Since 7 is the sum, the other side is 7 - x.

Step 3 :

Also product of the side lengths = 10
x(7 - x) = 10
7x - x$^2$ = 10
or x$^2$ - 7x + 10 = 0 (Solve using factoring method)
x$^2$ - 5x - 2x + 10 = 0
x(x - 5) - 2(x - 5) = 0
(x - 2)(x - 5) = 0

Step 4 :

x - 2 = 0 and x - 5 = 0
x = 2 and x = 5
If one side is 2 units then second side is 5 units or vice versa.