Solve -x-5+4x^2

$$- x - 5 + 4 x ^ { 2 }$$

Factor

$$\left(4x-5\right)\left(x+1\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$4x^{2}-x-5$$

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-1 ab=4\left(-5\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

Rewrite 4x^{2}-x-5 as \left(4x^{2}-5x\right)+\left(4x-5\right).

$$\left(4x^{2}-5x\right)+\left(4x-5\right)$$

Factor out x in 4x^{2}-5x.

$$x\left(4x-5\right)+4x-5$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(x+1\right)$$

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.

$$4x^{2}-x-5=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 4\left(-5\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-1\right)±\sqrt{1-16\left(-5\right)}}{2\times 4}$$

Multiply -16 times -5.

$$x=\frac{-\left(-1\right)±\sqrt{1+80}}{2\times 4}$$

Add 1 to 80.

$$x=\frac{-\left(-1\right)±\sqrt{81}}{2\times 4}$$

Take the square root of 81.

$$x=\frac{-\left(-1\right)±9}{2\times 4}$$

The opposite of -1 is 1.

$$x=\frac{1±9}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{1±9}{8}$$

Now solve the equation x=\frac{1±9}{8} when ± is plus. Add 1 to 9.

$$x=\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{1±9}{8} when ± is minus. Subtract 9 from 1.

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

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{5}{4} for x_{1} and -1 for x_{2}.

$$4x^{2}-x-5=4\left(x-\frac{5}{4}\right)\left(x-\left(-1\right)\right)$$

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

$$4x^{2}-x-5=4\left(x-\frac{5}{4}\right)\left(x+1\right)$$

Subtract \frac{5}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-x-5=4\times \frac{4x-5}{4}\left(x+1\right)$$

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

$$4x^{2}-x-5=\left(4x-5\right)\left(x+1\right)$$

Evaluate

$$\left(4x-5\right)\left(x+1\right)$$