Question

How do you factor \[{{x}^{3}}+8{{x}^{2}}+15x\]?

Answer 1

Hint: We will solve this polynomial first by using hit and trial method to get one factor of \[x\], for which the expression gives a value zero. Then we will use that factor to divide the given polynomial and we will get a quadratic equation. Solving this obtained quadratic equation, we will get two values of \[x\]. Therefore, in total we will get three values of \[x\] for which the expression gives a value zero.

Complete step by step solution:
According to the given question, we have given a polynomial with the degree 3. Therefore, it is safe to assume that we will get three values of \[x\]. To start solving the question, we will first have to find a factor using the hit and trial method with which we will eventually divide the expression.
So we will take up values like 1, -1, 2, -2…and so on.
If we see the expression, the arithmetic operator used is addition, so in order to get a value of zero from the expression, we need to have a negative sign as well, so the value of \[x\] is some negative number.
Let’s start with \[x=-1\], we get,
\[{{(-1)}^{3}}+8{{(-1)}^{2}}+15(-1)=-8\] not the required result.
When \[x=-2\], we have
\[{{(-2)}^{3}}+8{{(-2)}^{2}}+15(-2)=-6\] not the required result.
And when \[x=-3\], we get
\[{{(-3)}^{3}}+8{{(-3)}^{2}}+15(-3)=-72+72=0\], required result
That is, \[x=-3\] is a factor of the given expression, so we will divide the given expression with \[(x+3)\].
We have,
\[(x+3)\overset{{{x}^{2}}+5x}{\overline{\left){\begin{align}
  & {{x}^{3}}+8{{x}^{2}}+15x \\
 & \underline{-({{x}^{3}}+3{{x}^{2}})} \\
 & 0{{x}^{3}}+5{{x}^{2}}+15x \\
 & \underline{-(0{{x}^{3}}+5{{x}^{2}}+15x)} \\
 & 0 \\
\end{align}}\right.}}\]
The quotient we got is a quadratic equation, we will solve it to obtain the other two values of \[x\].
On solving we get,
\[{{x}^{2}}+5x=0\]
Taking \[x\] common we get,
\[\Rightarrow x(x+5)=0\]
\[x=0,x+5=0\]
\[x=0,x=-5\]
Therefore, we have the three values of \[x\] for the expression given to us, they are:
\[x=0,x=-5,x=-3\]
That is, for these values of the expression gives the value zero.

Note:
The hit and trial method used in the beginning should always start with the least possible value that \[x\] can take, and gradually be increased. This ensures that all possible entities are checked for the factor of the given expression. The division of the expression with the factor should be carried out neatly to avoid extra terms getting added or subtracted.