Solve 6x^4-19x^3-32x^2+52x-16=0

$$6 x ^ { 4 } - 19 x ^ { 3 } - 32 x ^ { 2 } + 52 x - 16 = 0$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=4$$

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

$$x=-2$$

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 -16 and q divides the leading coefficient 6. List all candidates \frac{p}{q}.

$$±\frac{8}{3},±\frac{16}{3},±8,±16,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1,±\frac{1}{6},±\frac{1}{2}$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{4}-19x^{3}-32x^{2}+52x-16 by x+2 to get 6x^{3}-31x^{2}+30x-8. Solve the equation where the result equals to 0.

$$6x^{3}-31x^{2}+30x-8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8 and q divides the leading coefficient 6. List all candidates \frac{p}{q}.

$$±\frac{4}{3},±\frac{8}{3},±4,±8,±\frac{2}{3},±2,±\frac{1}{3},±1,±\frac{1}{6},±\frac{1}{2}$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{3}-31x^{2}+30x-8 by x-4 to get 6x^{2}-7x+2. Solve the equation where the result equals to 0.

$$6x^{2}-7x+2=0$$

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 6 for a, -7 for b, and 2 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 6\times 2}}{2\times 6}$$

Do the calculations.

$$x=\frac{7±1}{12}$$

Solve the equation 6x^{2}-7x+2=0 when ± is plus and when ± is minus.

$$x=\frac{1}{2} x=\frac{2}{3}$$

List all found solutions.

$$x=-2 x=4 x=\frac{1}{2} x=\frac{2}{3}$$