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

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

Factor

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

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.

$$a+b=-5 ab=2\left(-3\right)=-6$$

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 -5.

$$a=-6 b=1$$

Rewrite 2x^{2}-5x-3 as \left(2x^{2}-6x\right)+\left(x-3\right).

$$\left(2x^{2}-6x\right)+\left(x-3\right)$$

Factor out 2x in 2x^{2}-6x.

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

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

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

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.

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

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.

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 2}$$

Add 25 to 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$x=\frac{-\left(-5\right)±7}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±7}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±7}{4}$$

Now solve the equation x=\frac{5±7}{4} when ± is plus. Add 5 to 7.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{5±7}{4} when ± is minus. Subtract 7 from 5.

$$x=-\frac{2}{4}$$

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

$$x=-\frac{1}{2}$$

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

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

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

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

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Evaluate

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