Question

p(x) =kx^3 + 3x^2 - 3 and q(x) =2x^3 - 5x + k, when divided by x-4 leave the same remainder in each case, find the value of k. ​

Answer 1

$\large{\red{\underline{\underline{\bold{Given:-}}}}}$

• p(x)= kx³+3x²-3 and q(x)=2x³-5x+k when divided by x-4 leaves same remainder

$\large{\blue{\underline{\underline{\bold{To Find:-}}}}}$

• The value of when p(x) and f(x) is divided by x-4

$\large{\green{\underline{\underline{\bold{Solution:-}}}}}$

$\rightarrow$ {As we know that we should use remainder theorem.Remainder theorem says that when x-4 then the value of x is 4.So substitute x in place of p(x) and f(x)}

$\rightarrow$ p(x)= kx³+3x²-3 , q(x)= 2x³-5x+k

$\rightarrow$ p(4)= k(4)³+3(4)²-3,q(4)=2(4)³-5(4)+k

$\rightarrow$ p(4)= 64k+48-3 , q(4)= 128-20+k

$\rightarrow$ p(4)= 64k+45 , q(4)= 108+k

• Now equate from both equations

$\longrightarrow$ 64k+45..............(i)

$\longrightarrow$ k+108.................(ii)

$\rightarrow$ 64k+45= k+108

$\rightarrow$ 64k-k= 108-45

$\rightarrow$ 63k= 63

$\rightarrow$ k= 1