Solve 8x^4+28x^3+20x^2

$$8 x ^ { 4 } + 28 x ^ { 3 } + 20 x ^ { 2 }$$

Factor

$$4\left(x+1\right)\left(2x+5\right)x^{2}$$

Solution Steps

Factor out 4.

$$4\left(2x^{4}+7x^{3}+5x^{2}\right)$$

Consider 2x^{4}+7x^{3}+5x^{2}. Factor out x^{2}.

$$x^{2}\left(2x^{2}+7x+5\right)$$

Consider 2x^{2}+7x+5. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\times 5=10$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 7.

$$a=2 b=5$$

Rewrite 2x^{2}+7x+5 as \left(2x^{2}+2x\right)+\left(5x+5\right).

$$\left(2x^{2}+2x\right)+\left(5x+5\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(x+1\right)+5\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(2x+5\right)$$

Rewrite the complete factored expression.

$$4x^{2}\left(x+1\right)\left(2x+5\right)$$

Evaluate

$$4\left(x+1\right)\left(2x+5\right)x^{2}$$