Solve 4x^4-36x^3+72x^2

$$4 x ^ { 4 } - 36 x ^ { 3 } + 72 x ^ { 2 }$$

Factor

$$4\left(x-6\right)\left(x-3\right)x^{2}$$

Solution Steps

Factor out 4.

$$4\left(x^{4}-9x^{3}+18x^{2}\right)$$

Consider x^{4}-9x^{3}+18x^{2}. Factor out x^{2}.

$$x^{2}\left(x^{2}-9x+18\right)$$

Consider x^{2}-9x+18. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+18. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 18=18$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -9.

$$a=-6 b=-3$$

Rewrite x^{2}-9x+18 as \left(x^{2}-6x\right)+\left(-3x+18\right).

$$\left(x^{2}-6x\right)+\left(-3x+18\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-6\right)-3\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x-3\right)$$

Rewrite the complete factored expression.

$$4x^{2}\left(x-6\right)\left(x-3\right)$$

Evaluate

$$4\left(x-6\right)\left(x-3\right)x^{2}$$