Solve 2x^{6}+x^5-9x^4+12x^3-14x^2+11x-3

$$2 x ^ { 6 } + x ^ { 5 } - 9 x ^ { 4 } + 12 x ^ { 3 } - 14 x ^ { 2 } + 11 x - 3$$

Factor

$$\left(2x-1\right)\left(x+3\right)\left(x-1\right)^{2}\left(x^{2}+1\right)$$

Solution Steps

To factor the expression, solve the equation where it equals to 0.

$$2x^{6}+x^{5}-9x^{4}+12x^{3}-14x^{2}+11x-3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -3 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{6}+x^{5}-9x^{4}+12x^{3}-14x^{2}+11x-3 by x-1 to get 2x^{5}+3x^{4}-6x^{3}+6x^{2}-8x+3. To factor the result, solve the equation where it equals to 0.

$$2x^{5}+3x^{4}-6x^{3}+6x^{2}-8x+3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 3 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{5}+3x^{4}-6x^{3}+6x^{2}-8x+3 by x-1 to get 2x^{4}+5x^{3}-x^{2}+5x-3. To factor the result, solve the equation where it equals to 0.

$$2x^{4}+5x^{3}-x^{2}+5x-3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -3 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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}+5x^{3}-x^{2}+5x-3 by x+3 to get 2x^{3}-x^{2}+2x-1. To factor the result, solve the equation where it equals to 0.

$$2x^{3}-x^{2}+2x-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

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}+2x-1 by 2\left(x-\frac{1}{2}\right)=2x-1 to get x^{2}+1. To factor the result, solve the equation where it equals to 0.

$$x^{2}+1=0$$

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 1 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-4}}{2}$$

Polynomial x^{2}+1 is not factored since it does not have any rational roots.

$$x^{2}+1$$

Rewrite the factored expression using the obtained roots.

$$\left(2x-1\right)\left(x+3\right)\left(x-1\right)^{2}\left(x^{2}+1\right)$$

Evaluate

$$\left(2x-1\right)\left(x+3\right)\left(x-1\right)^{2}\left(x^{2}+1\right)$$