Answer :
We have to factorize the polynomial:
[tex]2x^3+10x^2-28x[/tex]using the gratest common factor (GFC).
Then, we have to find all x-intercepts of the polynomial.
We can factorize each term as:
[tex]\begin{gathered} 2x^3=2\cdot x^3 \\ 10x^2=2\cdot5\cdot x^2 \\ 28x=2\cdot2\cdot7\cdot x=2^2\cdot7\cdot x \end{gathered}[/tex]We have 2 and x as common factors of the polynomial, so we write:
[tex]\begin{gathered} 2x^3+10x^2-28x \\ 2x(x^2)+2x(5x)-2x(2\cdot7) \\ 2x(x^2+5x-14) \end{gathered}[/tex]We now need to apply the quadratic formula to find the roots of the quadratic polynomial in parenthesis:
[tex]\begin{gathered} x^2+5x-14 \\ \Rightarrow x=\frac{-5\pm\sqrt[]{5^2-4\cdot1\cdot(-14)}}{2\cdot1} \\ x=\frac{-5\pm\sqrt[]{25+56}}{2} \\ x=\frac{-5\pm\sqrt[]{81}}{2} \\ x=\frac{-5\pm9}{2} \\ x_1=\frac{-5-9}{2}=\frac{-14}{2}=-7 \\ x_2=\frac{-5+9}{2}=\frac{4}{2}=2 \end{gathered}[/tex]We can now factorize the polynomial as:
[tex]x(x^2+5x-14)=x(x+7)(x-2)[/tex]This factorized form gives us the x-intercepts:
[tex]\begin{gathered} f(x)=0 \\ x(x+7)(x-2)=0 \\ \Rightarrow x_1=0 \\ x_2=-7 \\ x_3=2 \end{gathered}[/tex]Answer:
The factorized polynomial is x(x+7)(x-2).
The x-intercepts are x = 0, x = -7 and x =2.
