Solve 10x^2-50x-x+5

$$10 x ^ { 2 } - 50 x - x + 5$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$10x^{2}-51x+5$$

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-51 ab=10\times 5=50$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 50.

$$-1,-50 -2,-25 -5,-10$$

Calculate the sum for each pair.

$$-1-50=-51 -2-25=-27 -5-10=-15$$

The solution is the pair that gives sum -51.

$$a=-50 b=-1$$

Rewrite 10x^{2}-51x+5 as \left(10x^{2}-50x\right)+\left(-x+5\right).

$$\left(10x^{2}-50x\right)+\left(-x+5\right)$$

Factor out 10x in the first and -1 in the second group.

$$10x\left(x-5\right)-\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

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

Evaluate

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

Solution Steps

Combine -50x and -x to get -51x.

$$10x^{2}-51x+5$$