Solve 5x^2+19x-4=

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

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.

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 5

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

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{19}{5} rs = -\frac{4}{5}$$

Two numbers r and s sum up to -\frac{19}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{19}{5} = -\frac{19}{10}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{19}{10} - u s = -\frac{19}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{5}

$$(-\frac{19}{10} - u) (-\frac{19}{10} + u) = -\frac{4}{5}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{361}{100} - u^2 = -\frac{4}{5}$$

Simplify the expression by subtracting \frac{361}{100} on both sides

$$-u^2 = -\frac{4}{5}-\frac{361}{100} = -\frac{441}{100}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{441}{100} u = \pm\sqrt{\frac{441}{100}} = \pm \frac{21}{10}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{19}{10} - \frac{21}{10} = -4 s = -\frac{19}{10} + \frac{21}{10} = 0.200$$

Evaluate

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