Name: Application of the Remainder Theorem: Example 2
Uploaded: 2019-07-19
Duration: 4 min 59 s
Description: What are the conditions to be a factor of a polynomial? Can you guess what the condition for x+1 to be a Factor of a Polynomial is? Watch the video to learn...

Question

Without actual division, prove that

x^{4} + 2 x^{3} - 2 x^{2} + 2 x - 3

is exactly divisible by

x^{2} + 2 x - 3

Accepted Answer

If $p (x)$ is a polynomial and $α$ is a real number, then $x - α$ is a factor of $p (x)$ iff $p (α) = 0$ .

Let $f (x) = x^{4} + 2 x^{3} - 2 x^{2} + 2 x - 3$ and $g (x) = x^{2} + 2 x - 3$ .

We have, $g (x) = x^{2} + 2 x - 3 = x^{2} + 3 x - x - 3 = x (x + 3) - 1 (x + 3) = (x + 3) (x - 1)$ .

So, $f (x) = x^{4} + 2 x^{3} - 2 x^{2} + 2 x - 3$ is exactly divisible by $g (x) = x^{2} + 2 x - 3$ if both $(x + 3)$ and $(x - 1)$ are factors of $f (x) = x^{4} + 2 x^{3} - 2 x^{2} + 2 x - 3$ or if both $f (- 3) = 0$ and $f (1) = 0$ .

Now, we have, $f (- 3) = {(- 3)}^{4} + 2 \times {(- 3)}^{3} - 2 \times {(- 3)}^{2} + 2 \times (- 3) - 3 = 81 + 2 \times (- 27) - 2 \times 9 - 6 - 3$

$= 81 - 54 - 18 - 6 - 3 = 0$

And, $f (1) = (- 1)^{4} + 2 \times (1)^{3} - 2 \times (1)^{2} + 2 \times (1) - 3 = 1 + 2 \times 1 - 2 \times 1 + 2 - 3$

$= 1 + 2 - 2 + 2 - 3 = 0$

So, we obtained, $f (- 3) = 0 = f (1)$ .

Therefore, $(x + 3)$ and $(x - 1)$ are factors of $x^{4} + 2 x^{3} - 2 x^{2} + 2 x - 3$ and thus, we can say, $x^{4} + 2 x^{3} - 2 x^{2} + 2 x - 3$ is exactly divisible by $x^{2} + 2 x - 3$ .

Hence, proved.

Without actual division, prove that $x^{4} + 2 x^{3} - 2 x^{2} + 2 x - 3$ is exactly divisible by $x^{2} + 2 x - 3$

Important Questions on Factorisation

Learn and Prepare for any exam you want !

Without actual division, prove that x4+2x3-2x2+2x-3 is exactly divisible by x2+2x-3

Important Questions on Factorisation

Learn and Prepare for any exam you want !

Without actual division, prove that $x^{4} + 2 x^{3} - 2 x^{2} + 2 x - 3$ is exactly divisible by $x^{2} + 2 x - 3$