Question

# If the remainder on division of ${x^3} + 2{x^2} + kx + 3$ by $x - 3$ is 21, find the quotient and value of k. Hence, find the zeros of the cubic polynomial ${x^3} + 2{x^2} + kx - 18$.

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Hint: We will use the remainder formula to find k from the given equation and long division process to find the quotient. We will also use the division formula of polynomials to get the zeros of the polynomial.

Given that the remainder on division of ${x^3} + 2{x^2} + kx + 3$ by $x - 3$ is 21
We have the following terms:
Dividend: $f(x) = {x^3} + 2{x^2} + kx + 3$
Divisor: ${\text{ }}g\left( x \right){\text{ }} = {\text{ }}x{\text{ }} - {\text{ }}3$ and remainder, $r{\text{ }}\left( x \right){\text{ }} = {\text{ }}21$
Using the remainder formula, we have the following expression:
$f\left( 3 \right) = 21$
$\Rightarrow {(3)^3} + 2.{(3)^2} + k.(3) + 3 = 21 \\ \Rightarrow 27 + 18 + 3k + 3 = 21 \\ \Rightarrow 3k = - 27 \\ \Rightarrow k = - 9 \\$
So, the polynomial is, $p(x) = {x^3} + 2{x^2} - 9x + 3$
Now, from the long division, we get,
${x^3}\; + {\text{ }}2{x^2}\; - {\text{ }}9x{\text{ }} + {\text{ }}3{\text{ }} = {\text{ }}\left( {x{\text{ }} - {\text{ }}3{\text{ }}} \right){\text{ }}({x^2}\; + {\text{ }}5x{\text{ }} + {\text{ }}6){\text{ }} + {\text{ }}21$
∴ The quotient $= {\text{ }}{x^2}\; + {\text{ }}5x{\text{ }} + {\text{ }}6$
Clearly, ${x^3}\; + {\text{ }}2{x^2}\; - {\text{ }}9x{\text{ }}-21 + 3 =$ ${x^3}\; + {\text{ }}2{x^2}\; - {\text{ }}9x{\text{ }}-{\text{ }}18$ is divisible by, $x - 3$
$= {x^3} - 3{x^2} + 5{x^2} - 15x + 6x - 18 \\ = {x^2}(x - 3) + 5x(x - 3) + 6(x - 3) \\$
$= {\text{ }}\left( {x{\text{ }} - {\text{ }}3{\text{ }}} \right){\text{ }}\left( {{x^2}\; + {\text{ }}5x{\text{ }} + {\text{ }}6} \right)\;$
On further splitting of middle terms we get,
$= {\text{ }}\left( {x{\text{ }} - {\text{ }}3{\text{ }}} \right){\text{ }}\left( {{x^2} + 3x + 2x + {\text{ }}6} \right)$
On further simplification we get,
$= {\text{ }}\left( {x{\text{ }} - {\text{ }}3{\text{ }}} \right){\text{ }}\left( {x{\text{ }} + {\text{ }}2} \right)\left( {x{\text{ }} + {\text{ }}3} \right)$
For, now, $(x - 3)$we have, $x = 3$
Then, for, ($x + 2$) we have, $x = - 2$
And also, for, ($x + 3$) we have, $x = - 3$
Therefore, the zeroes of ${x^3}\; + {\text{ }}2{x^2}\; - {\text{ }}9x\; - {\text{ }}18$ are 3, -2 and -3.

Note: We have the remainder theorem as , $f(x) = g(x).h(x) + r(x)$. Where $f(x)$is the dividend and $g(x)$is the divisor. We also have $r\left( x \right)$as the reminder. This type of problems are built with the concept of long division altogether.