Solve 16x^4-32x^3+48x^2-32x+7=0

$$16 { x }^{ 4 } -32 { x }^{ 3 } +48 { x }^{ 2 } -32x+7=0$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{6}i}{2}\approx 0.5+1.224744871i$$

$$x=\frac{-\sqrt{6}i+1}{2}\approx 0.5-1.224744871i$$

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

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

$$±\frac{7}{16},±\frac{7}{8},±\frac{7}{4},±\frac{7}{2},±7,±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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 16x^{4}-32x^{3}+48x^{2}-32x+7 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 8x^{3}-12x^{2}+18x-7. Solve the equation where the result equals to 0.

$$8x^{3}-12x^{2}+18x-7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -7 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

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

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 8x^{3}-12x^{2}+18x-7 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 4x^{2}-4x+7. Solve the equation where the result equals to 0.

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

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 4 for a, -4 for b, and 7 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±\sqrt{-96}}{8}$$

Solve the equation 4x^{2}-4x+7=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{6}i+1}{2} x=\frac{1+\sqrt{6}i}{2}$$

List all found solutions.

$$x=\frac{1}{2} x=\frac{-\sqrt{6}i+1}{2} x=\frac{1+\sqrt{6}i}{2}$$

Solve for x

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

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

$$±\frac{7}{16},±\frac{7}{8},±\frac{7}{4},±\frac{7}{2},±7,±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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 16x^{4}-32x^{3}+48x^{2}-32x+7 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 8x^{3}-12x^{2}+18x-7. Solve the equation where the result equals to 0.

$$8x^{3}-12x^{2}+18x-7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -7 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

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

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 8x^{3}-12x^{2}+18x-7 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 4x^{2}-4x+7. Solve the equation where the result equals to 0.

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

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 4 for a, -4 for b, and 7 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±\sqrt{-96}}{8}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

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