Name: Factor Theorem: Illustration
Uploaded: 2019-07-19
Duration: 4 min 17 s
Description: The factorisation of a cubic polynomial might look difficult. Let us learn how to find factors of a cubic polynomial with an illustration. Watch the video fo...

Question

Write the second degree polynomial $x^{2} - 20 x + 91$ as the product of two first degree polynomials.

Accepted Answer

The given second degree polynomial is $x^{2} - 20 x + 91$ .

We need to write the given second degree polynomial as the product of two first degree polynomials.

If $x - a$ is a factor of $p (x)$ , then $p (a) = 0$ . In other words, $a$ must be a solution of the equation $x^{2} - 20 x + 91 = 0$

By, comparing $x^{2} - 20 x + 91 = 0$ with $a x^{2} + b x + c = 0$ , we get $a = 1$ , $b = - 20$ and $c = 91$

As we have, formula to solve the quadratic equation is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Thus, $x = \frac{- (- 20) \pm \sqrt{{(- 20)}^{2} - 4 \times 1 \times 91}}{2 \times 1}$

$\Rightarrow x = \frac{20 \pm \sqrt{400 - 364}}{2} = \frac{20 \pm \sqrt{36}}{2}$

$\Rightarrow x = \frac{20 \pm 6}{2}$

$\Rightarrow x = \frac{20 + 6}{2}, \frac{20 - 6}{2}$

$\Rightarrow x = 13, 7$

This means $p (13) = 0$ and $p (7) = 0$

From this, it follows that $x - 13$ and $x - 7$ are factors of $p (x)$ .

Therefore, $x^{2} - 20 x + 91 = (x - 13) (x - 7)$ .

Write the second degree polynomial $x^{2} - 20 x + 91$ as the product of two first degree polynomials.

Important Questions on Polynomials

Learn and Prepare for any exam you want !

Write the second degree polynomial x2-20x+91 as the product of two first degree polynomials.

Important Questions on Polynomials

Learn and Prepare for any exam you want !

Write the second degree polynomial $x^{2} - 20 x + 91$ as the product of two first degree polynomials.