Solve 2x^2-x-3

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

Factor

$$\left(2x-3\right)\left(x+1\right)$$

Solution Steps

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

$$a+b=-1 ab=2\left(-3\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

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

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

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

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(x+1\right)$$

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.

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

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\times 2}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

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

The opposite of -1 is 1.

$$x=\frac{1±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±5}{4}$$

Now solve the equation x=\frac{1±5}{4} when ± is plus. Add 1 to 5.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{1±5}{4} when ± is minus. Subtract 5 from 1.

$$x=-\frac{4}{4}$$

Divide -4 by 4.

$$x=-1$$

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

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

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

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

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 2, the greatest common factor in 2 and 2.

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

Evaluate

$$\left(2x-3\right)\left(x+1\right)$$