Divide $p (x)$ by $g (x)$
$p (x) = x^{3} - 4 x^{2} + x + 6, g (x) = x - 3$

Open in App

Solution

$p (x) = x^{3} - 4 x^{2} + x + 6 ⟹ p (3) = 27 - 36 + 3 + 6 = 0$
$⟹ p (x) = (x - 3) (x^{2} + x - 2)$
$\frac{p (x)}{g (x)} = \frac{x^{3} - 4 x^{2} + x + 6}{x - 3} = x^{2} + x - 2$

Suggest Corrections

Similar questions

Q. Divide

p (x)

g (x)

in the following case and verify division algorithm.

p (x) = x^{3} + 4 x^{2} - 5 x + 6; g (x) = x + 1

Q. p(x) =

x^{3} + 4 x^{2} - 5 x + 6

g(x) = x + 1
and verify with

p (x) [g (x) \times q (x)] + r (x)

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x)=2 $x^{3}$ + $x^{2}$ −2x−1,g(x)=x+1
(ii) p(x)= $x^{3}$ +3 $x^{2}$ +3x+1,g(x)=x+2
(iii) p(x)= $x^{3}$ −4 $x^{2}$ +x+6,g(x)=x−3

Q. Find the quotient and remainder on dividing

p (x)

g (x)

in the following case, without actual division.

p (x) = x^{3} + 4 x^{2} - 6 x + 2; g (x) = x - 3

Q. Question 2
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

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

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

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

Join BYJU'S Learning Program

Divide p(x) by g(x)p(x)=x3−4x2+x+6,g(x)=x−3

p(x)=x3−4x2+x+6⟹p(3)=27−36+3+6=0⟹p(x)=(x−3)(x2+x−2)p(x)g(x)=x3−4x2+x+6x−3=x2+x−2

Divide $p (x)$ by $g (x)$
$p (x) = x^{3} - 4 x^{2} + x + 6, g (x) = x - 3$

$p (x) = x^{3} - 4 x^{2} + x + 6 ⟹ p (3) = 27 - 36 + 3 + 6 = 0$
$⟹ p (x) = (x - 3) (x^{2} + x - 2)$
$\frac{p (x)}{g (x)} = \frac{x^{3} - 4 x^{2} + x + 6}{x - 3} = x^{2} + x - 2$