Given $p (x) = x^{3} + 8 x^{2} - 7 x + 12$ and $g (x) = x - 1$ . Find the remainder when $p (x)$ is divided by $g (x)$ .

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Solution

The correct option is B 14
Given:
$p (x) = x^{3} + 8 x^{2} - 7 x + 12$ is divided by $g (x) = x - 1$

Let the remainder be $r (x)$ . Using Remainder theorem, remainder is given by $p (1)$ .
$r (x) = p (1)$
$r (x) = 1^{3} + 8 \times 1^{2} - 7 \times 1 + 12 = 14$

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Similar questions

$p (x) = x^{3} + 8 x^{2} - 7 x + 12$ and $g (x) = x - 1$ . If $p (x)$ is divided by $g (x)$ , it gives $q (x)$ and $r (x)$ as quotient and remainder respectively. If $a$ is the degree of $q (x)$ and $b$ is the degree of $r (x)$ , $(a - b) = ?$ .

Q. By Remainder Theorem find the remainder, when

p (x)

is divided by

g (x)

, where

p (x) = x^{3} - 2 x^{2} - 4 x - 1

g (x) = x + 1

Q. Question 14
By Remainder theorem, find the remainder when p(x) is divided by g(x).

(i)

p (x) = x^{3} - 2 x^{2} - 4 x - 1, g (x) = x + 1

(ii)

p (x) = x^{3} - 3 x^{2} + 4 x + 50, g (x) = x - 3

(iii)

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3, g (x) = 2 x - 1

(iv)

p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where $(p x) = x^{3} - 6 x^{2} + 9 x + 3, g (x) = x - 1$

p(x) = $x^{3}$ + 8 $x^{2}$ - 7x + 12 and g(x) = x - 1. If p(x) is divided by g(x), it gives q(x) and r(x) as quotient and remainder respectively. If 'a' is the degree of q(x) and b is the degree of r(x), what is the value of (a - b)?

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Given p(x)=x3+8x2−7x+12 and g(x)=x−1. Find the remainder when p(x) is divided by g(x).

The correct option is B 14Given: p(x)=x3+8x2−7x+12 is divided by g(x)=x−1 Let the remainder be r(x). Using Remainder theorem, remainder is given by p(1). r(x)=p(1) r(x)=13+8×12−7×1+12=14

Given $p (x) = x^{3} + 8 x^{2} - 7 x + 12$ and $g (x) = x - 1$ . Find the remainder when $p (x)$ is divided by $g (x)$ .

The correct option is B 14
Given:
$p (x) = x^{3} + 8 x^{2} - 7 x + 12$ is divided by $g (x) = x - 1$

Let the remainder be $r (x)$ . Using Remainder theorem, remainder is given by $p (1)$ .
$r (x) = p (1)$
$r (x) = 1^{3} + 8 \times 1^{2} - 7 \times 1 + 12 = 14$