Solve 4x^2+2x-5-3x

$$4 { x }^{ 2 } +2x-5-3x$$

Factor

$$\left(4x-5\right)\left(x+1\right)$$

Solution Steps

Multiply and combine like terms.

$$4x^{2}-x-5$$

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-1 ab=4\left(-5\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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.

$$a=-5 b=4$$

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

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

Factor out x in 4x^{2}-5x.

$$x\left(4x-5\right)+4x-5$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(x+1\right)$$

Evaluate

$$\left(4x-5\right)\left(x+1\right)$$

Solution Steps

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

$$4x^{2}-x-5$$