Question

Find all the real zeros of the polynomial. Use the quadratic formula if necessary.

$P\left(x\right)=5x^3+7x^2-4x-6$

Answer 1

Finding zeroes of the polynomial:

Using the hit and trial method let's check for $x=-1$

$$\begin{gathered} P(-1)=-5+7+4-6 \\ P(-1)=0 \end{gathered}$$

so, $x=-1$ is one of the root of cubic polynomial, now on dividing the $P\left(x\right)=5x^3+7x^2-4x-6$ by the root factor $x+1$ we get,

$\begin{array}{rcl}5x^3+7x^2-4x-6& =& 5x^3+5x^2+2x^2+2x-6x-6\\ & =& 5x^2(x+1)+2x(x+1)-6(x+1)\\ & =& (x+1)(5x^2+2x-6)\end{array}$

Another factor is $5x^2+2x-6$.

Now, finding roots of this quadratic equation.

Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where, $a=5,b=2,c=-6$ we get,

$$\begin{gathered} x=\frac{-2\pm \sqrt{2^2+4.5.6}}{2.5} \\ \Rightarrow x=\frac{-2\pm \sqrt{124}}{10} \\ \Rightarrow x=\frac{-2\pm 2\sqrt{31}}{10} \\ \Rightarrow x=\frac{-1+\sqrt{31}}{5},\frac{-1-\sqrt{31}}{5} \end{gathered}$$

Hence, the roots of the polynomial are $x=-1,\frac{-1+\sqrt{31}}{5},\frac{-1-\sqrt{31}}{5}$.