Solve x*2x+x*5-3*2x-3*5

$$x \times 2x+x \times 5-3 \times 2x-3 \times 5$$

Evaluate

$$\left(x-3\right)\left(2x+5\right)$$

Solution Steps

Multiply x and x to get x^{2}.

$$x^{2}\times 2+x\times 5-3\times 2x-3\times 5$$

Multiply 3 and 2 to get 6.

$$x^{2}\times 2+x\times 5-6x-3\times 5$$

Combine x\times 5 and -6x to get -x.

$$x^{2}\times 2-x-3\times 5$$

Multiply 3 and 5 to get 15.

$$x^{2}\times 2-x-15$$

Factor

$$\left(x-3\right)\left(2x+5\right)$$

Solution Steps

Multiply and combine like terms.

$$2x^{2}-x-15$$

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-15\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

Rewrite 2x^{2}-x-15 as \left(2x^{2}-6x\right)+\left(5x-15\right).

$$\left(2x^{2}-6x\right)+\left(5x-15\right)$$

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

$$2x\left(x-3\right)+5\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(2x+5\right)$$