Solve x^2-5x+6quad7x^3

$$x ^ { 2 } - 5 x + 6 \quad 7 x ^ { 3 }$$

Factor

$$x\left(3x-1\right)\left(14x+5\right)$$

Solution Steps

Factor out x.

$$x\left(x-5+42x^{2}\right)$$

Consider x-5+42x^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$42x^{2}+x-5$$

Factor the expression by grouping. First, the expression needs to be rewritten as 42x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=1 ab=42\left(-5\right)=-210$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -210.

$$-1,210 -2,105 -3,70 -5,42 -6,35 -7,30 -10,21 -14,15$$

Calculate the sum for each pair.

$$-1+210=209 -2+105=103 -3+70=67 -5+42=37 -6+35=29 -7+30=23 -10+21=11 -14+15=1$$

The solution is the pair that gives sum 1.

$$a=-14 b=15$$

Rewrite 42x^{2}+x-5 as \left(42x^{2}-14x\right)+\left(15x-5\right).

$$\left(42x^{2}-14x\right)+\left(15x-5\right)$$

Factor out 14x in the first and 5 in the second group.

$$14x\left(3x-1\right)+5\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(14x+5\right)$$

Rewrite the complete factored expression.

$$x\left(3x-1\right)\left(14x+5\right)$$

Evaluate

$$x\left(3x-1\right)\left(14x+5\right)$$

Solution Steps

Multiply 6 and 7 to get 42.

$$x^{2}-5x+42x^{3}$$