Solve -5x^2-x+4

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

Factor

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

Solution Steps

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

$$a+b=-1 ab=-5\times 4=-20$$

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.

$$a=4 b=-5$$

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

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

Factor out -x in the first and -1 in the second group.

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

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

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

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.

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

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.

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

Multiply -4 times -5.

$$x=\frac{-\left(-1\right)±\sqrt{1+20\times 4}}{2\left(-5\right)}$$

Multiply 20 times 4.

$$x=\frac{-\left(-1\right)±\sqrt{1+80}}{2\left(-5\right)}$$

Add 1 to 80.

$$x=\frac{-\left(-1\right)±\sqrt{81}}{2\left(-5\right)}$$

Take the square root of 81.

$$x=\frac{-\left(-1\right)±9}{2\left(-5\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±9}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{1±9}{-10}$$

Now solve the equation x=\frac{1±9}{-10} when ± is plus. Add 1 to 9.

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

Divide 10 by -10.

$$x=-1$$

Now solve the equation x=\frac{1±9}{-10} when ± is minus. Subtract 9 from 1.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and \frac{4}{5} for x_{2}.

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

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

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

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

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

Cancel out 5, the greatest common factor in -5 and 5.

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

Evaluate

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