Solve 2x^2-7x+-4

$$2 x ^ { 2 } - 7 x + - 4$$

Factor

$$\left(x-4\right)\left(2x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\left(-4\right)=-8$$

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 -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -7.

$$a=-8 b=1$$

Rewrite 2x^{2}-7x-4 as \left(2x^{2}-8x\right)+\left(x-4\right).

$$\left(2x^{2}-8x\right)+\left(x-4\right)$$

Factor out 2x in 2x^{2}-8x.

$$2x\left(x-4\right)+x-4$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(2x+1\right)$$

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.

$$2x^{2}-7x-4=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-4\right)}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-4\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$x=\frac{-\left(-7\right)±\sqrt{49+32}}{2\times 2}$$

Add 49 to 32.

$$x=\frac{-\left(-7\right)±\sqrt{81}}{2\times 2}$$

Take the square root of 81.

$$x=\frac{-\left(-7\right)±9}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±9}{4}$$

Now solve the equation x=\frac{7±9}{4} when ± is plus. Add 7 to 9.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{7±9}{4} when ± is minus. Subtract 9 from 7.

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

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 4 for x_{1} and -\frac{1}{2} for x_{2}.

$$2x^{2}-7x-4=2\left(x-4\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$2x^{2}-7x-4=2\left(x-4\right)\left(x+\frac{1}{2}\right)$$

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

$$2x^{2}-7x-4=2\left(x-4\right)\times \frac{2x+1}{2}$$

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

$$2x^{2}-7x-4=\left(x-4\right)\left(2x+1\right)$$

Evaluate

$$\left(x-4\right)\left(2x+1\right)$$