Solve 4x^2+3x-7+(-2x^2)-2x+(-3)

$$4 x ^ { 2 } + 3 x - 7 + ( - 2 x ^ { 2 } ) - 2 x + ( - 3 )$$

Evaluate

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

Solution Steps

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}+3x-7-2x-3$$

Combine 3x and -2x to get x.

$$2x^{2}+x-7-3$$

Subtract 3 from -7 to get -10.

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

Factor

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

Solution Steps

Multiply and combine like terms.

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

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

$$a+b=1 ab=2\left(-10\right)=-20$$

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 -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite 2x^{2}+x-10 as \left(2x^{2}-4x\right)+\left(5x-10\right).

$$\left(2x^{2}-4x\right)+\left(5x-10\right)$$

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

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

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

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