Solve 10^2-2x+5x-1-10x^2=

$${ 10 }^{ 2 } -2x+5x-1-10 { x }^{ 2 } =$$

Evaluate

$$-\left(10x-33\right)\left(x+3\right)$$

Solution Steps

Calculate 10 to the power of 2 and get 100.

$$100-2x+5x-1-10x^{2}$$

Combine -2x and 5x to get 3x.

$$100+3x-1-10x^{2}$$

Subtract 1 from 100 to get 99.

$$99+3x-10x^{2}$$

Factor

$$-\left(10x-33\right)\left(x+3\right)$$

Solution Steps

Multiply and combine like terms.

$$-10x^{2}+3x+99$$

Factor the expression by grouping. First, the expression needs to be rewritten as -10x^{2}+ax+bx+99. To find a and b, set up a system to be solved.

$$a+b=3 ab=-10\times 99=-990$$

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 -990.

$$-1,990 -2,495 -3,330 -5,198 -6,165 -9,110 -10,99 -11,90 -15,66 -18,55 -22,45 -30,33$$

Calculate the sum for each pair.

$$-1+990=989 -2+495=493 -3+330=327 -5+198=193 -6+165=159 -9+110=101 -10+99=89 -11+90=79 -15+66=51 -18+55=37 -22+45=23 -30+33=3$$

The solution is the pair that gives sum 3.

$$a=33 b=-30$$

Rewrite -10x^{2}+3x+99 as \left(-10x^{2}+33x\right)+\left(-30x+99\right).

$$\left(-10x^{2}+33x\right)+\left(-30x+99\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(10x-33\right)-3\left(10x-33\right)$$

Factor out common term 10x-33 by using distributive property.

$$\left(10x-33\right)\left(-x-3\right)$$