Solve 2x^3=7x^2-x-10

$$2 { x }^{ 3 } =7 { x }^{ 2 } -x-10$$

Solve for x

$$x=2$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x=-1$$

Solution Steps

Subtract 7x^{2} from both sides.

$$2x^{3}-7x^{2}=-x-10$$

Add x to both sides.

$$2x^{3}-7x^{2}+x=-10$$

Add 10 to both sides.

$$2x^{3}-7x^{2}+x+10=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 10 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±5,±10,±\frac{5}{2},±1,±2,±\frac{1}{2}$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-7x^{2}+x+10 by x+1 to get 2x^{2}-9x+10. Solve the equation where the result equals to 0.

$$2x^{2}-9x+10=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, -9 for b, and 10 for c in the quadratic formula.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\times 10}}{2\times 2}$$

Do the calculations.

$$x=\frac{9±1}{4}$$

Solve the equation 2x^{2}-9x+10=0 when ± is plus and when ± is minus.

$$x=2 x=\frac{5}{2}$$

List all found solutions.

$$x=-1 x=2 x=\frac{5}{2}$$