Question

If $x+k$ is the GCD of ${{x}^{2}}-2x-15$ and ${{x}^{3}}+27$, find the value of $k$.

Answer 1

Hint: In the problem we have the GCD of the two expressions. So, we will factorize the given expressions and calculate the Greatest Common Divisor of both the expressions and equate it to the given GCD $x+k$ to find the value of $k$.

Complete step-by-step answer:
Given that, $x+k$ is the GCD of ${{x}^{2}}-2x-15$ and ${{x}^{3}}+27$.
Consider the expression ${{x}^{2}}-2x-15$ which is a quadratic equation. To factorize the quadratic equation $a{{x}^{2}}+bx+c$we will split the middle term $bx={{b}_{1}}x+{{b}_{2}}x$ where ${{b}_{1}}.{{b}_{2}}=c$. So to factorize the quadratic expression ${{x}^{2}}-2x-15$, we will split the middle term $-2x=3x-5x$ since $\left( 3 \right)\left( -5 \right)=-15$
${{x}^{2}}-2x-15={{x}^{2}}+3x-5x-15$
Taking $x$ common from the terms ${{x}^{2}}+3x$ and $-5$ from the terms $-5x-15$, then we will get
$\Rightarrow {{x}^{2}}-2x-15=x\left( x+3 \right)-5\left( x+3 \right)$
Now taking $\left( x+3 \right)$ common from the above expression, then we will get
$\Rightarrow {{x}^{2}}-2x-15=\left( x+3 \right)\left( x-5 \right)$
$\therefore $ The factors of the expression ${{x}^{2}}-2x-15$ are $x+3$ and $x-5$.
Now consider the expression ${{x}^{3}}+27$. We can write $27={{3}^{3}}$, then we will get
${{x}^{3}}+27={{x}^{3}}+{{3}^{3}}$
We have the arithmetic expression ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$, then we will get
$\begin{align}
  & \Rightarrow {{x}^{3}}+27=\left( x+3 \right)\left( {{x}^{2}}-3x+{{3}^{2}} \right) \\
 & \Rightarrow {{x}^{3}}+27=\left( x+3 \right)\left( {{x}^{2}}-3x+9 \right) \\
\end{align}$
$\therefore $ The factors of the expression ${{x}^{3}}+27$ are $x+3$ and ${{x}^{2}}-3x+9$.
From the factors of the given expressions the GCD of the expressions ${{x}^{2}}-2x-15$, ${{x}^{3}}+27$ is $x+3$.
But in the problem, we have given that the GCD is $x+k$. Comparing the given GCD to calculated GCD we will get the value of $k$ is equal to $3$.

Note: We can find the factors of quadratic equation by train and error method. In this method we will substitute $x=\pm 1,\pm 2,\pm 3,...$ and find the values of $x$ where the given expression gives zero value. Let us say for values $\alpha $ and $\beta $ the expression given zero values then we can write the factors of the given expression as $\left( x-\alpha \right)\left( x-\beta \right)$. We can extend this method to cubic equation also, but here we will get three values say $\alpha $,$\beta $ ,$\gamma $ and the factors of the cubic expression as $\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)$.