Solve 4x^2-3x-7

$$4 { x }^{ 2 } -3x-7$$

Factor

$$\left(4x-7\right)\left(x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=-3 ab=4\left(-7\right)=-28$$

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 -28.

$$1,-28 2,-14 4,-7$$

Calculate the sum for each pair.

$$1-28=-27 2-14=-12 4-7=-3$$

The solution is the pair that gives sum -3.

$$a=-7 b=4$$

Rewrite 4x^{2}-3x-7 as \left(4x^{2}-7x\right)+\left(4x-7\right).

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

Factor out x in 4x^{2}-7x.

$$x\left(4x-7\right)+4x-7$$

Factor out common term 4x-7 by using distributive property.

$$\left(4x-7\right)\left(x+1\right)$$

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.

$$4x^{2}-3x-7=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 4\left(-7\right)}}{2\times 4}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 4\left(-7\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9-16\left(-7\right)}}{2\times 4}$$

Multiply -16 times -7.

$$x=\frac{-\left(-3\right)±\sqrt{9+112}}{2\times 4}$$

Add 9 to 112.

$$x=\frac{-\left(-3\right)±\sqrt{121}}{2\times 4}$$

Take the square root of 121.

$$x=\frac{-\left(-3\right)±11}{2\times 4}$$

The opposite of -3 is 3.

$$x=\frac{3±11}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{3±11}{8}$$

Now solve the equation x=\frac{3±11}{8} when ± is plus. Add 3 to 11.

$$x=\frac{14}{8}$$

Reduce the fraction \frac{14}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{4}$$

Now solve the equation x=\frac{3±11}{8} when ± is minus. Subtract 11 from 3.

$$x=-\frac{8}{8}$$

Divide -8 by 8.

$$x=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{4} for x_{1} and -1 for x_{2}.

$$4x^{2}-3x-7=4\left(x-\frac{7}{4}\right)\left(x-\left(-1\right)\right)$$

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

$$4x^{2}-3x-7=4\left(x-\frac{7}{4}\right)\left(x+1\right)$$

Subtract \frac{7}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-3x-7=4\times \frac{4x-7}{4}\left(x+1\right)$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4x^{2}-3x-7=\left(4x-7\right)\left(x+1\right)$$

Evaluate

$$\left(4x-7\right)\left(x+1\right)$$