Given that
We have an equation as
[tex]y=2x^3+x^2-7x-6[/tex]
Explanation -
Here we have to find the values of x when y = 0.
So we will use HINT AND TRIAL METHOD.
In this, we will substitute such values of x that make y = 0.
So first we will put x = -1
[tex]\begin{gathered} y=2(-1)^3+(1)^2-7(-1)-6 \\ y=-2+1+7-6=8-8=0 \end{gathered}[/tex]
So the first point will be (-1,0)
Now we will substitute x = 2
[tex]\begin{gathered} y=2(2)^3+2^2-7\times2-6 \\ y=16+4-14-6=20-20 \\ y=0 \end{gathered}[/tex]
So the second point is (2,0)
Now we will substitute x = -3/2
[tex]\begin{gathered} y=2\times(-\frac{3}{2})^3+(-\frac{3}{2})^2-7(-\frac{3}{2})-6 \\ y=\frac{2\times-27}{8}+\frac{9}{4}+\frac{21}{2}-6 \\ y=-\frac{54}{8}+\frac{9}{4}+\frac{21}{2}-6=-\frac{27}{4}+\frac{9}{4}+\frac{21}{2}-6 \\ y=-\frac{18}{4}+\frac{21}{2}-6=-\frac{9}{2}+\frac{21}{2}-6 \\ y=\frac{12}{2}-6=6-6=0 \end{gathered}[/tex]
So the third point is (-3/2,0)
Hence option A is correct.
And the final answer is (-1,0), (2,0), (-3/2,0)
