Solve 20x^2-2x+50x-5

$$20 x ^ { 2 } - 2 x + 50 x - 5$$

Factor

$$\left(10x-1\right)\left(2x+5\right)$$

Solution Steps

Multiply and combine like terms.

$$20x^{2}+48x-5$$

Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=48 ab=20\left(-5\right)=-100$$

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 -100.

$$-1,100 -2,50 -4,25 -5,20 -10,10$$

Calculate the sum for each pair.

$$-1+100=99 -2+50=48 -4+25=21 -5+20=15 -10+10=0$$

The solution is the pair that gives sum 48.

$$a=-2 b=50$$

Rewrite 20x^{2}+48x-5 as \left(20x^{2}-2x\right)+\left(50x-5\right).

$$\left(20x^{2}-2x\right)+\left(50x-5\right)$$

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

$$2x\left(10x-1\right)+5\left(10x-1\right)$$

Factor out common term 10x-1 by using distributive property.

$$\left(10x-1\right)\left(2x+5\right)$$

Evaluate

$$\left(10x-1\right)\left(2x+5\right)$$

Solution Steps

Combine -2x and 50x to get 48x.

$$20x^{2}+48x-5$$