Solve 12x^4-13x^3-35x^2

$$12 x ^ { 4 } - 13 x ^ { 3 } - 35 x ^ { 2 }$$

Factor

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

Solution Steps

Factor out x^{2}.

$$x^{2}\left(12x^{2}-13x-35\right)$$

Consider 12x^{2}-13x-35. Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=-13 ab=12\left(-35\right)=-420$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -420.

$$1,-420 2,-210 3,-140 4,-105 5,-84 6,-70 7,-60 10,-42 12,-35 14,-30 15,-28 20,-21$$

Calculate the sum for each pair.

$$1-420=-419 2-210=-208 3-140=-137 4-105=-101 5-84=-79 6-70=-64 7-60=-53 10-42=-32 12-35=-23 14-30=-16 15-28=-13 20-21=-1$$

The solution is the pair that gives sum -13.

$$a=-28 b=15$$

Rewrite 12x^{2}-13x-35 as \left(12x^{2}-28x\right)+\left(15x-35\right).

$$\left(12x^{2}-28x\right)+\left(15x-35\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(3x-7\right)+5\left(3x-7\right)$$

Factor out common term 3x-7 by using distributive property.

$$\left(3x-7\right)\left(4x+5\right)$$

Rewrite the complete factored expression.

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

Evaluate

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