Determine the value of $p$ if $(x + 3)$ is a factor of $x^{3} - 3 x^{2} - p x + 24$ .

Open in App

Solution

According to factor theorem, when $f (x)$ is divided by $(x + 3), f (- 3) = 0$
$f (x) = x^{3} - 3 x^{2} - p x + 24$
$f (- 3) = (- 3)^{3} - 3 (- 3)^{2} - p (- 3) + 24$
$⟹ 0 = - 27 - 27 + 3 p + 24$
$⟹ 0 = 3 p - 30$
$⟹ 3 p = 30$
$⟹ p = \frac{30}{3} = 10$

Suggest Corrections

Similar questions

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. 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

x+1 is a factor of p(x)= x³+3x²+3x+1 , so what is the value of p(-1)

Q. Prove that

(x - 2), (x + 3)

and

(x - 4)

are factors of the polynomial:

p (x) = x^{3} - 3 x^{2} - 10 x + 24

Q. If

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

and

p (\frac{\sqrt{3}}{2}) = 0

then by factor theorem the corresponding factor of

p (x)

Join BYJU'S Learning Program

Determine the value of p if (x+3) is a factor of x3−3x2−px+24.

According to factor theorem, when f(x) is divided by (x+3),f(−3)=0f(x)=x3−3x2−px+24f(−3)=(−3)3−3(−3)2−p(−3)+24⟹0=−27−27+3p+24⟹0=3p−30⟹3p=30⟹p=303=10

Determine the value of $p$ if $(x + 3)$ is a factor of $x^{3} - 3 x^{2} - p x + 24$ .

According to factor theorem, when $f (x)$ is divided by $(x + 3), f (- 3) = 0$
$f (x) = x^{3} - 3 x^{2} - p x + 24$
$f (- 3) = (- 3)^{3} - 3 (- 3)^{2} - p (- 3) + 24$
$⟹ 0 = - 27 - 27 + 3 p + 24$
$⟹ 0 = 3 p - 30$
$⟹ 3 p = 30$
$⟹ p = \frac{30}{3} = 10$