Question

How do you solve $15{{x}^{3}}-65{{x}^{2}}-150x=0$?

Answer 1

Take 5x common and form a quadratic expression inside the bracket. Now, use the middle term split method to factorize the obtained quadratic expression. Substitute each factor equal to 0 one – by – one to get the values of x. The three values of x that will be obtained will be our answer.

Complete step by step answer:
Here, we have been provided with the polynomial expression: $15{{x}^{3}}-65{{x}^{2}}-150x=0$ and we have been asked to solve it. That means we have to find the values of x.
Now, as we can see that we have a cubic polynomial in which the constant term is 0. So, taking 5x common from all the terms, we get,
$\Rightarrow 5x\left( 3{{x}^{2}}-13x-30 \right)=0$
Dividing both the sides with 5, we get,
$\Rightarrow x\left( 3{{x}^{2}}-13x-30 \right)=0$
Now, we have a product of a linear and a quadratic expression, so now we need to find the factored form of $3{{x}^{2}}-13x-30$. Let us use the middle term split method to do this.
Here, we need to break -13x into two terms such that their product equals the product of the constant term (-30) and $+3{{x}^{2}}$, i.e., $-90{{x}^{2}}$. So, writing 90 as the product of its prime factors, we get,
$\Rightarrow 90=2\times 3\times 3\times 5$
So, grouping -18x and 5x, we have,
$\Rightarrow -18x+5x=-13x$ and $\left( -18x \right)\times \left( 5x \right)=-90{{x}^{2}}$
Therefore, the polynomial expression can be simplified as: -

& \Rightarrow x\left( 3{{x}^{2}}-18x+5x-30 \right)=0 \\\ & \Rightarrow x\left[ 3x\left( x-6 \right)+5\left( x-6 \right) \right]=0 \\\ \end{aligned}$$ Taking $$\left( x-6 \right)$$ common we get, $$\Rightarrow x\left( x-6 \right)\left( 3x+5 \right)=0$$ Substituting each term equal to 0, we get, $$\Rightarrow x=0$$ or $$\left( x-6 \right)=0$$ or $$\left( 3x+5 \right)=0$$. $$\Rightarrow x=0$$ or $$x=-6$$ or $$x=\dfrac{-5}{3}$$ Hence, the three roots of the given cubic equation can be given as: - $$x\in \left\\{ 0,-6,\dfrac{-5}{3} \right\\}$$. **Note:** One may note that this was a cubic equation and that is why we obtained three roots. So, the highest power of the variable determines that number of factors and roots of the given polynomial. You may see that in the provided expression the constant term is 0. If the polynomial were of the form $$a{{x}^{3}}+b{{x}^{2}}+cx+d=0$$, where (a, b, c, d) $$\ne $$ 0 then in that case we have to find one factor by hit and trial method and the remaining two by the middle term split method.