Solve x^5-4x^4+3x^3-8x^2+32x-24

$$x ^ { 5 } - 4 x ^ { 4 } + 3 x ^ { 3 } - 8 x ^ { 2 } + 32 x - 24$$

Factor

$$\left(x-3\right)\left(x-2\right)\left(x-1\right)\left(x^{2}+2x+4\right)$$

Solution Steps

Do the grouping x^{5}-4x^{4}+3x^{3}-8x^{2}+32x-24=\left(x^{5}-4x^{4}+3x^{3}\right)+\left(-8x^{2}+32x-24\right), and factor out x^{3} in the first and -8 in the second group.

$$x^{3}\left(x^{2}-4x+3\right)-8\left(x^{2}-4x+3\right)$$

Factor out common term x^{2}-4x+3 by using distributive property.

$$\left(x^{2}-4x+3\right)\left(x^{3}-8\right)$$

Consider x^{2}-4x+3. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

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

Rewrite x^{2}-4x+3 as \left(x^{2}-3x\right)+\left(-x+3\right).

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

Factor out x in the first and -1 in the second group.

$$x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x-1\right)$$

Consider x^{3}-8. Rewrite x^{3}-8 as x^{3}-2^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

$$\left(x-2\right)\left(x^{2}+2x+4\right)$$

Rewrite the complete factored expression. Polynomial x^{2}+2x+4 is not factored since it does not have any rational roots.

$$\left(x-3\right)\left(x-2\right)\left(x-1\right)\left(x^{2}+2x+4\right)$$

Evaluate

$$\left(x-3\right)\left(x-1\right)\left(x^{3}-8\right)$$