Question

Find the value of ‘p’ for which the polynomial 2x^4 + 3x^3 + 2px^2 +3x + 6 is exactly
divisible by (x+2)

Answer 1

p(x)=2x⁴ + 3x³ + 2px² +3x + 6----->1
x+2=0
x=-2
put x=-2 in equation 1.
p(-2)=2(-2)⁴+3(-2)³+2p(-2)²+3(-2)+6
=>32-24+8p-6+6
=>8+8p=0
=>p=-8/8
=>p=-1

Answer 2

Answer:

$p = - 1$

Step-by-step explanation:

To find the value of ‘p’ for which the polynomial

$2 {x}^{4} + 3 {x}^{3} + 2p {x}^{2} + 3x + 6$

is exactly divisible by (x+2)

Apply Remainder theorem: put the value of x=-2 in the polynomial

$2 {( - 2)}^{4} + 3 {( - 2)}^{3} + 2p {( - 2)}^{2} + 3( - 2) + 6 = 0 \\ \\ 32 - 24 + 8p - 6 + 6 = 0 \\ \\ 8p = - 32 + 24 \\ \\ 8p = - 8 \\ \\ p = \frac{ - 8}{8} \\ \\ p = - 1$

Thus, value of p = -1

Hope it helps you.