what are x-intercepts for the equation y=2x^3-x^2+2x-1

Answer 1

To find the x-intercepts of the equation $y=2x^3-x^2+2x-1$, you need to set $y$ to 0 and solve for $x$:

$$2x^3-x^2+2x-1=0$$

This is a cubic equation, and finding its roots can be done through various methods such as graphing, synthetic division, or using the Rational Root Theorem.

1. Rational Root Theorem: Potential rational roots could be factors of the constant term (-1) divided by factors of the leading coefficient (2). The candidates are: $\pm 1,\pm \frac{1}{2}$.

2. Testing possible roots:

   • Let's test $x=1$: $$2(1{)}^3-(1{)}^2+2(1)-1=2-1+2-1=2(\text{not a root})$$
   • Now test $x=-1$: $$2(-1{)}^3-(-1{)}^2+2(-1)-1=-2-1-2-1=-6(\text{not a root})$$
   • Now test $x=\frac{1}{2}$: $$2{\left(\frac{1}{2}\right)}^3-{\left(\frac{1}{2}\right)}^2+2\left(\frac{1}{2}\right)-1=2\left(\frac{1}{8}\right)-\frac{1}{4}+1-1$$ $$=\frac{1}{4}-\frac{1}{4}+1-1=0(\text{is a root})$$

Since $x=\frac{1}{2}$ is a root, we can use synthetic division to divide the polynomial $2x^3-x^2+2x-1$ by $x-\frac{1}{2}$:

Synthetic Division:

0.5 |  2  -1   2  -1
    |      1   0.0  1.0
    ------------------
      2  0.0  2.0  0

The result is $2x^2+0x+2$.

Factoring the Remaining Quadratic:

We now solve the quadratic $2x^2+2=0$:

$$2x^2=-2\to x^2=-1\to x=i,-i$$

Conclusion:

The x-intercepts of the equation $y=2x^3-x^2+2x-1$ is $(\frac{1}{2},0)$ while $x=i$ and $x=-i$ are complex roots. Therefore, the only x-intercept in the real number set is:

$$\text{x-intercept: }(\frac{1}{2},0)$$