Solve 5x^2+10x-15

$$5 { x }^{ 2 } +10x-15$$

Factor

$$5\left(x-1\right)\left(x+3\right)$$

Solution Steps

Factor out 5.

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

Consider x^{2}+2x-3. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-1 b=3$$

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

$$\left(x^{2}-x\right)+\left(3x-3\right)$$

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

$$x\left(x-1\right)+3\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+3\right)$$

Rewrite the complete factored expression.

$$5\left(x-1\right)\left(x+3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$5x^{2}+10x-15=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-10±\sqrt{10^{2}-4\times 5\left(-15\right)}}{2\times 5}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 5\left(-15\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-10±\sqrt{100-20\left(-15\right)}}{2\times 5}$$

Multiply -20 times -15.

$$x=\frac{-10±\sqrt{100+300}}{2\times 5}$$

Add 100 to 300.

$$x=\frac{-10±\sqrt{400}}{2\times 5}$$

Take the square root of 400.

$$x=\frac{-10±20}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-10±20}{10}$$

Now solve the equation x=\frac{-10±20}{10} when ± is plus. Add -10 to 20.

$$x=\frac{10}{10}$$

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{-10±20}{10} when ± is minus. Subtract 20 from -10.

$$x=-\frac{30}{10}$$

Divide -30 by 10.

$$x=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -3 for x_{2}.

$$5x^{2}+10x-15=5\left(x-1\right)\left(x-\left(-3\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$5x^{2}+10x-15=5\left(x-1\right)\left(x+3\right)$$

Evaluate

$$5\left(x-1\right)\left(x+3\right)$$