Question

the polynomial p(x) = x³+ax²+bx-20 when divided by x-5 and x-3 leaves the remainder 0 and -2 respectively. Find the values of a and b.

Answer 1

Answer: it is in the below attachment....

Answer 2

Concept

The remainder theorem begins with a polynomial such as p (x). Where "p (x)" is the polynomial p whose variable is x. Then, according to the theorem, divide this polynomial p (x) by the linear factor x – a. Where a is just a number. Now we perform a long polynomial division leading to the polynomial q (x) (the variable "q" represents the "quotient polynomial"), and the remainder of the polynomial  is r (x). This can be expressed as:

   p (x) / x-a = q (x) + r (x)

Given

We have given a polynomial  $p(x)=x^3+ax^2+bx-20$  and which on dividing by $x-5$ and $x-3$ leaves remainder 0 and -2 respectively.

Find

We are asked to determine the values of a and b in the given polynomial

Solution

As $x-5$ and $x-3$ are factors of given polynomial p(x).

So the values of x will be following

$x-5=0\\x=5$         and     $x-3=0\\x=3$

Putting x = 5 in the given polynomial p(x) , we get

$p(5)=(5)^3+a(5)^2+b(5)-20$

Therefore, by applying the remainder theorem, p(5) should be equal to 0

$0=(5)^3+a(5)^2+b(5)-20\\\\20=125+25a+5b\\25a+5b=-105$.....(1)

Putting x = 3 in the given polynomial p(x) , we get

$p(3)=(3)^3+a(3)^2+b(3)-20$

Therefore, by applying the remainder theorem, p(3) should be equal to -2

$-2=(3)^3+a(3)^2+b(3)-20\\-2=27+9a+3b-20\\-2=7+9a+3b\\9a+3b=-9$....(2)

On solving equation(1) and (2) , we get

a = -9 , b = 24

Therefore, the values of a and b in the given polynomial are -9 and 24 respectively.

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