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

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

Evaluate

$$\left(x-2\right)\left(4x+1\right)$$

Solution Steps

Combine x^{2} and 3x^{2} to get 4x^{2}.

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

Combine -4x and -3x to get -7x.

$$4x^{2}-7x+3-5$$

Subtract 5 from 3 to get -2.

$$4x^{2}-7x-2$$

Factor

$$\left(x-2\right)\left(4x+1\right)$$

Solution Steps

Multiply and combine like terms.

$$4x^{2}-7x-2$$

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

$$a+b=-7 ab=4\left(-2\right)=-8$$

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 -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -7.

$$a=-8 b=1$$

Rewrite 4x^{2}-7x-2 as \left(4x^{2}-8x\right)+\left(x-2\right).

$$\left(4x^{2}-8x\right)+\left(x-2\right)$$

Factor out 4x in 4x^{2}-8x.

$$4x\left(x-2\right)+x-2$$

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

$$\left(x-2\right)\left(4x+1\right)$$