Solve (2x-7)(x^2-9)(2x+5)=91

$$( 2 x - 7 ) ( x ^ { 2 } - 9 ) ( 2 x + 5 ) = 91$$

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$x=4$$

$$x=\frac{1-\sqrt{65}}{4}\approx -1.765564437$$

$$x = \frac{\sqrt{65} + 1}{4} \approx 2.265564437$$

Solution Steps

Use the distributive property to multiply 2x-7 by x^{2}-9.

$$\left(2x^{3}-18x-7x^{2}+63\right)\left(2x+5\right)=91$$

Use the distributive property to multiply 2x^{3}-18x-7x^{2}+63 by 2x+5 and combine like terms.

$$4x^{4}-4x^{3}-71x^{2}+36x+315=91$$

Subtract 91 from both sides.

$$4x^{4}-4x^{3}-71x^{2}+36x+315-91=0$$

Subtract 91 from 315 to get 224.

$$4x^{4}-4x^{3}-71x^{2}+36x+224=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 224 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±56,±112,±224,±28,±14,±8,±16,±32,±7,±4,±\frac{7}{2},±2,±\frac{7}{4},±1,±\frac{1}{2},±\frac{1}{4}$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{4}-4x^{3}-71x^{2}+36x+224 by x-4 to get 4x^{3}+12x^{2}-23x-56. Solve the equation where the result equals to 0.

$$4x^{3}+12x^{2}-23x-56=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -56 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±14,±28,±56,±7,±\frac{7}{2},±2,±4,±8,±\frac{7}{4},±1,±\frac{1}{2},±\frac{1}{4}$$

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.

$$x=-\frac{7}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+12x^{2}-23x-56 by 2\left(x+\frac{7}{2}\right)=2x+7 to get 2x^{2}-x-8. Solve the equation where the result equals to 0.

$$2x^{2}-x-8=0$$

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, -1 for b, and -8 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{1±\sqrt{65}}{4}$$

Solve the equation 2x^{2}-x-8=0 when ± is plus and when ± is minus.

$$x=\frac{1-\sqrt{65}}{4} x=\frac{\sqrt{65}+1}{4}$$

List all found solutions.

$$x=4 x=-\frac{7}{2} x=\frac{1-\sqrt{65}}{4} x=\frac{\sqrt{65}+1}{4}$$