Question

Question number 9 solve it and send it please it is from remainder factor theorem

Answer 1

Let given polynomial,
p(x) =px³+4x²-3x+q=0
also given that given polynomial is exactly divisible by (x²-1).
Therefore (x²-1) is the factor of given polynomial.
(x²-1)=(x-1)(x+1)
Then (x-1) and (x+1) are also factor of given polynomial.
Then 1 and - 1 are zeroes of p(x).
p(x) =0

p(1)=0

p(1)³+4(1)²-3(1)+q=0

p+4-3+q=0

p+q=-1..................................................1
Now,
p(-1)=0

p(-1)³+4(-1)²-3(-1)+q=0

-p+4+3+q=0

-p+q=-7

p-q=7..................................................2
Adding eq1 and eq2 we get,
p+q+p-q=-1+7

2p=6

p=3
putting value of p in eq1 we get,
3+q=-1

q=-4
Hence value of p=3 and q=-4
Now the polynomial we have,
3x³+4x²-3x-4=0
Now,
=(3x³+4x²-3x-4)/(x²-1)

=3x+4
Now we have,
(3x³+4x²-3x-4)=(x²-1)(3x+4)

(3x³+4x²-3x-4)=(x-1)(x+1)(3x+4)

Hence all zeroes of given polynomial are 1,-1 and -4/3.