How do you determine the value of k if the remainder is 3 given $(kx^3+3x+1) div(x+2)$ ?

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Answer 1 · 2017-07-27T06:26:52

$k=-1$

The Remainder Theorem states that:

if the polynomial $P(x)" "$ is divided by $" "(x-a)" "$ the remainder is $P(a)$

In this case we have

$P(x)=kx^3+3x+1$

and the divisor $(x+2)=>a=-2$

$:.P(-2) = 3$

$:.P(-2)=k(-2)^3+3xx -2+1=3$

$=>-8k-6+1=3$

$-8k-5=3$

$-8k=8$

$=>k=-1$

How do you determine the value of k if the remainder is 3 given (kx^3+3x+1) div(x+2)(kx^3+3x+1) div(x+2)?