Solve 5x^2+19x-4

$$5 { x }^{ 2 } +19x-4$$

Factor

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

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.

$$a+b=19 ab=5\left(-4\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 19.

$$a=-1 b=20$$

Rewrite 5x^{2}+19x-4 as \left(5x^{2}-x\right)+\left(20x-4\right).

$$\left(5x^{2}-x\right)+\left(20x-4\right)$$

Factor out x in the first and 4 in the second group.

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

Factor out common term 5x-1 by using distributive property.

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

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.

$$5x^{2}+19x-4=0$$

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.

$$x=\frac{-19±\sqrt{19^{2}-4\times 5\left(-4\right)}}{2\times 5}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 5\left(-4\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-19±\sqrt{361-20\left(-4\right)}}{2\times 5}$$

Multiply -20 times -4.

$$x=\frac{-19±\sqrt{361+80}}{2\times 5}$$

Add 361 to 80.

$$x=\frac{-19±\sqrt{441}}{2\times 5}$$

Take the square root of 441.

$$x=\frac{-19±21}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-19±21}{10}$$

Now solve the equation x=\frac{-19±21}{10} when ± is plus. Add -19 to 21.

$$x=\frac{2}{10}$$

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

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-19±21}{10} when ± is minus. Subtract 21 from -19.

$$x=-\frac{40}{10}$$

Divide -40 by 10.

$$x=-4$$

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

$$5x^{2}+19x-4=5\left(x-\frac{1}{5}\right)\left(x-\left(-4\right)\right)$$

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

$$5x^{2}+19x-4=5\left(x-\frac{1}{5}\right)\left(x+4\right)$$

Subtract \frac{1}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}+19x-4=5\times \frac{5x-1}{5}\left(x+4\right)$$

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

$$5x^{2}+19x-4=\left(5x-1\right)\left(x+4\right)$$

Evaluate

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