How do you determine the value of k if the remainder is 3 given $(x^{3} + 4 x^{2} - x + k) \div (x - 1)$ $(x^3+4x^2-x+k)div(x-1)$ ?

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Answer 1 · 2016-12-16T01:54:40

$k = - 1$ $k = -1$ .

the remainder theorem states that if a polynomial $p (x)$ $p(x)$ is divided by $x - a$ $x - a$ , then the remainder is given by $p (a)$ $p(a)$ .

Here, $a = 1$ $a = 1$ and the remainder is $3$ $3$ .

So:

${(1)}^{3} + 4 {(1)}^{2} - 1 + k = 3$ $(1)^3 + 4(1)^2 - 1 + k = 3$

$1 + 4 - 1 + k = 3$ $1 + 4 - 1 + k = 3$

$k = - 1$ $k = -1$

Hopefully this helps!

How do you determine the value of k if the remainder is 3 given (x^3+4x^2-x+k)div(x-1)(x3+4x2−x+k)÷(x−1)(x^3+4x^2-x+k)div(x-1)?