Question

P(x)=2x³-3px²+px+q have x-1 as a factor and have -52 as reminder when divide by x+2..
What value of p and q?

Answer 1

Let x-1=0
x=1

P(x)=2x^3-3px^2+px+q
P(1)=2(1)^3-3p(1)^2+p(1)+q
=2-3p+p+q

Let x+2=0
x=-2

P(x)=2x^3-3px^2+px+q
P(-2)=2(-2)^3-3p(-2)^2+p(-2)+q
=-16-12p-2p+q=54
=16-12p-2p+q-54

Please mark as brainliest answer

Answer 2

Answer:

p=17/6 and q=11/3

Step-by-step explanation:

Given:

P(x)=2x³-3px²+px+q

As x-1 is a factor of the given polynomial it must satisfies the polynomial.

x-1=0

x=1

substitute x=1 in the given polynomial.

2(1)³-3p(1)²+p(1)+q=0

2-3p+p+q=0

2=2p-q............................... equation (1)

According to the given condition -52 is a remainder when divided by x+2.

Now,

x+2=0

x=-2

substitute x=-2 in the given polynomial and simplify.

-52=2(-2)³-3p(-2)²+p(-2)+q

-52=-16-12p-2p+q

-52+16=-14p+q

-36=-14p+q...................................equation (2)

on solving equation (1) and (2) we get

p=17/6 and q=11/3.