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

If

(- p x^{2} + q x + 6 + x^{3})

has

(x - 1)

as a factor and leaves a remainder

4

when divided by

(x + 1)

, then find the values of

p

and

q

.

Accepted Answer

Given polynomial is $- p x^{2} + q x + 6 + x^{3}$ .

Let $f (x) = x^{3} - p x^{2} + q x + 6$ .

According to the question $(x - 1)$ is a factor, so $f (x) = 0$ will be satisfied.

Therefore, $f (1) = (x^{3} - p x^{2} + q x + 6)$

$\Rightarrow (1^{3} - p {(1)}^{2} + q (1) + 6) = 0$

$\Rightarrow 1 + q - p + 6 = 0$

$\Rightarrow q = p - 7 . . . . . . . . . . . . . (i)$

Also, when divided by $(x + 1)$ , leaves $4$ as a remainder, i.e $f (- 1) = 4$ ,

Therefore, $f (- 1) = (x^{3} - p x^{2} + q x + 6)$

$\Rightarrow ({(- 1)}^{3} - p {(- 1)}^{2} + q (- 1) + 6) = 4$

$\Rightarrow - 1 - p - q + 6 = 4$

$\Rightarrow p + q - 1 = 0 . . . . . . . . . . . . (ii)$

Substitute $(i)$ in $(ii)$ , we get,

$p + (p - 7) - 1 = 0$

$\Rightarrow 2 p = 8$

$∴ p = 4$

Substitute $p = 4$ in the Eq. $(i)$ , we get,

$q = p - 7 = 4 - 7$

$∴ q = - 3$

Hence, the values are $p = 4$ and $q = - 3$ .

If $(- p x^{2} + q x + 6 + x^{3})$ has $(x - 1)$ as a factor and leaves a remainder $4$ when divided by $(x + 1)$ , then find the values of $p$ and $q$ .

Important Questions on Factorisation of Polynomials

Learn and Prepare for any exam you want !

If -px2+qx+6+x3 has x-1 as a factor and leaves a remainder 4 when divided byx+1, then find the values of p and q.

Important Questions on Factorisation of Polynomials

Learn and Prepare for any exam you want !

If $(- p x^{2} + q x + 6 + x^{3})$ has $(x - 1)$ as a factor and leaves a remainder $4$ when divided by $(x + 1)$ , then find the values of $p$ and $q$ .