The remainder when $2x^3 + 9x^2 + 7x + 3$ is divided by x - k is 9, how do you find k?

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The remainder when $2x^3 + 9x^2 + 7x + 3$ is divided by x - k is 9, how do you find k?

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Answer 1 · 2015-10-18T19:47:53

The remainder of dividing $f(x) = 2x^3+9x^2+7x+3$ by $(x-k)$ is $f(k)$ , so solve $f(k) = 9$ using the rational root theorem and factoring to find:

$k = 1/2, -2$ or $-3$

Explanation:

If you attempt to divide $f(x) = 2x^3+9x^2+7x+3$ by $x-k$ you end up with a remainder of $f(k)$ ...

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So if the remainder is $9$ , we are basically trying to solve $f(k) = 9$

$2k^3+9k^2+7k+3 = 9$

Subtract $9$ from both sides to get:

$2k^3+9k^2+7k-6 = 0$

By the rational root theorem, any rational roots of this cubic will be of the form $p/q$ in lowest terms, where $p, q in ZZ$ , $q != 0$ , $p$ a divisor of the constant term $-6$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the possible rational roots are:

$+-1/2$ , $+-1$ , $+-3/2$ , $+-2$ , $+-3$ , $+-6$

Let's try the first one:

$f(1/2) = 1/4+9/4+7/2-6 = (1+9+14-24)/4 = 0$

so $k = 1/2$ is a root and $(2k-1)$ is a factor.

Divide by $(2k-1)$ to find:

$2k^3+9k^2+7k-6 = (2k-1)(k^2+5k+6) = (2k-1)(k+2)(k+3)$

So the possible solutions are:

$k = 1/2$ , $k = -2$ and $k = -3$

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The remainder when $2x^3 + 9x^2 + 7x + 3$ is divided by x - k is 9, how do you find k?

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Explanation:

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The remainder when $2x^3 + 9x^2 + 7x + 3$ is divided by x - k is 9, how do you find k?