Solve 2x^4-7x^3-7x^2+35x-15=0

$$2 x ^ { 4 } - 7 x ^ { 3 } - 7 x ^ { 2 } + 35 x - 15 = 0$$

Solve for x

$$x=\sqrt{5}\approx 2.236067977$$

$$x=-\sqrt{5}\approx -2.236067977$$

$$x=3$$

$$x=\frac{1}{2}=0.5$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -15 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{15}{2},±15,±\frac{5}{2},±5,±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{4}-7x^{3}-7x^{2}+35x-15 by x-3 to get 2x^{3}-x^{2}-10x+5. Solve the equation where the result equals to 0.

$$2x^{3}-x^{2}-10x+5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 5 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{5}{2},±5,±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-x^{2}-10x+5 by 2\left(x-\frac{1}{2}\right)=2x-1 to get x^{2}-5. Solve the equation where the result equals to 0.

$$x^{2}-5=0$$

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 1 for a, 0 for b, and -5 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-5\right)}}{2}$$

Do the calculations.

$$x=\frac{0±2\sqrt{5}}{2}$$

Solve the equation x^{2}-5=0 when ± is plus and when ± is minus.

$$x=-\sqrt{5} x=\sqrt{5}$$

List all found solutions.

$$x=3 x=\frac{1}{2} x=-\sqrt{5} x=\sqrt{5}$$