Question

The value of 'p' for which the polynomial x3+4x2-px+8 is exactly divisible by (x-2) is

Answer 1

$P(x)=x^3+4x^2-px+8\ is divisible\ exactly\ by\ (x-2) \\ \\x = 2\ \ is\ a\ zero\ of\ P(x).\\ \\Hence\ P(2)=0\\ \\2^3+4*2^2-p*2+8=0\\ \\8+16-2p+8=0\\ \\2p=32\\ \\p=16$

Answer 2

Answer:

The value of p is 16

Solution:

Given the polynomial $f(x)= x^3+4x^2-px+8$

For f(x) to be exactly divisible by 2, we must have $f(2)=0$ by remainder and factor theorem

Since given that, (x – 2) is exactly divisible by 2.

Thus, x = 2

Substituting x=2 in the given expression, we get,

$\begin{aligned} f ( 2 ) = & 2 ^ { 3 } + 4 \times 2 ^ { 2 } - 2 p + 8 = 0 \\\\ & 8 + 16 - 2 p + 8 = 0 \\\\ & 16 + 16 - 2 p = 0 \end{aligned}$

$\begin{array} { c } { 32 - 2 p = 0 } \\\\ { 2 p = 32 } \\\\ { p = 16 } \end{array}$

Which gives p=16  

Thus, the value of p will be 16.