Question

The polynomial p(x) = 2x3+kx2-3x +5 and q(x) = x3+ 2x2–x + k, when divided by (x – 2) leave the remainders r1 and r2.resp. find the value of k, if r1 = r2

Answer 1

Main concept.

Remainder theorem

The remainder theorem can find out the remainder of the division by a linear polynomial through substitution, without actual division.

Solution.

$p(x)=2x^3+kx^2-3x+5$ and $q(x)=x^3+2x^2-x+k$ are divided by $x-2$. By remainder theorem $r_{1}=16+4k-6+5=4k+15$ and $r_{2}=8+8-2+k=k+14$.

$\implies 4k+15=k+14$

$\implies 3k=-1$

$\implies \boxed{k=-\dfrac{1}{3} }$

Answer 2

Step-by-step explanation:

Given:-

p(x)=2x³+kx²-3x+5. remainder=r1

q(x)=x³+2x²-x+k. remainder=r2

Find:-

k=? if r1=r2

solution:-

p(2)=16+4k²-6+5=r1

q(2)=8+8-2+k=r2

as r1=r2=r(say)

4k²-15=r

14-k=r

4k²-15=14-k

4k²+k-29=0

$k = (\sqrt{465} - 1) \div 8$