Solve 2/7+13/35x-6/5x^2

$$\frac{ 2 }{ 7 } + \frac{ 13 }{ 35 } x- \frac{ 6 }{ 5 } { x }^{ 2 }$$

Factor

$$\frac{\left(-14x-5\right)\left(3x-2\right)}{35}$$

Solution Steps

Factor out \frac{1}{35}.

$$\frac{10+13x-42x^{2}}{35}$$

Consider 10+13x-42x^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-42x^{2}+13x+10$$

Factor the expression by grouping. First, the expression needs to be rewritten as -42x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=13 ab=-42\times 10=-420$$

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 -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 -42x^{2}+13x+10 as \left(-42x^{2}+28x\right)+\left(-15x+10\right).

$$\left(-42x^{2}+28x\right)+\left(-15x+10\right)$$

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

$$-14x\left(3x-2\right)-5\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(-14x-5\right)$$

Rewrite the complete factored expression.

$$\frac{\left(3x-2\right)\left(-14x-5\right)}{35}$$

Evaluate

$$-\frac{6x^{2}}{5}+\frac{13x}{35}+\frac{2}{7}$$