Solve a^4+4a^3+2a^2-4a-3

$$a ^ { 4 } + 4 a ^ { 3 } + 2 a ^ { 2 } - 4 a - 3$$

Factor

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

Solution Steps

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

$$a^{4}+4a^{3}+2a^{2}-4a-3=0$$

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 1. List all candidates \frac{p}{q}.

$$±3,±1$$

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.

$$a=1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{4}+4a^{3}+2a^{2}-4a-3 by a-1 to get a^{3}+5a^{2}+7a+3. To factor the result, solve the equation where it equals to 0.

$$a^{3}+5a^{2}+7a+3=0$$

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 1. List all candidates \frac{p}{q}.

$$±3,±1$$

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.

$$a=-1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{3}+5a^{2}+7a+3 by a+1 to get a^{2}+4a+3. To factor the result, solve the equation where it equals to 0.

$$a^{2}+4a+3=0$$

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, 4 for b, and 3 for c in the quadratic formula.

$$a=\frac{-4±\sqrt{4^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$a=\frac{-4±2}{2}$$

Solve the equation a^{2}+4a+3=0 when ± is plus and when ± is minus.

$$a=-3 a=-1$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

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