Solve 3x^2-2x-5

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

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-5\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite 3x^{2}-2x-5 as \left(3x^{2}-5x\right)+\left(3x-5\right).

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

Factor out x in 3x^{2}-5x.

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

Factor out common term 3x-5 by using distributive property.

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

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.

$$3x^{2}-2x-5=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$x=\frac{-\left(-2\right)±\sqrt{4+60}}{2\times 3}$$

Add 4 to 60.

$$x=\frac{-\left(-2\right)±\sqrt{64}}{2\times 3}$$

Take the square root of 64.

$$x=\frac{-\left(-2\right)±8}{2\times 3}$$

The opposite of -2 is 2.

$$x=\frac{2±8}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{2±8}{6}$$

Now solve the equation x=\frac{2±8}{6} when ± is plus. Add 2 to 8.

$$x=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{2±8}{6} when ± is minus. Subtract 8 from 2.

$$x=-\frac{6}{6}$$

Divide -6 by 6.

$$x=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{3} for x_{1} and -1 for x_{2}.

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

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

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

Subtract \frac{5}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 3, the greatest common factor in 3 and 3.

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

Evaluate

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