Solve 5p^2-48p-20

$$5 p ^ { 2 } - 48 p - 20$$

Factor

$$\left(p-10\right)\left(5p+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5p^{2}+ap+bp-20. To find a and b, set up a system to be solved.

$$a+b=-48 ab=5\left(-20\right)=-100$$

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 -100.

$$1,-100 2,-50 4,-25 5,-20 10,-10$$

Calculate the sum for each pair.

$$1-100=-99 2-50=-48 4-25=-21 5-20=-15 10-10=0$$

The solution is the pair that gives sum -48.

$$a=-50 b=2$$

Rewrite 5p^{2}-48p-20 as \left(5p^{2}-50p\right)+\left(2p-20\right).

$$\left(5p^{2}-50p\right)+\left(2p-20\right)$$

Factor out 5p in the first and 2 in the second group.

$$5p\left(p-10\right)+2\left(p-10\right)$$

Factor out common term p-10 by using distributive property.

$$\left(p-10\right)\left(5p+2\right)$$

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.

$$5p^{2}-48p-20=0$$

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.

$$p=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 5\left(-20\right)}}{2\times 5}$$

Square -48.

$$p=\frac{-\left(-48\right)±\sqrt{2304-4\times 5\left(-20\right)}}{2\times 5}$$

Multiply -4 times 5.

$$p=\frac{-\left(-48\right)±\sqrt{2304-20\left(-20\right)}}{2\times 5}$$

Multiply -20 times -20.

$$p=\frac{-\left(-48\right)±\sqrt{2304+400}}{2\times 5}$$

Add 2304 to 400.

$$p=\frac{-\left(-48\right)±\sqrt{2704}}{2\times 5}$$

Take the square root of 2704.

$$p=\frac{-\left(-48\right)±52}{2\times 5}$$

The opposite of -48 is 48.

$$p=\frac{48±52}{2\times 5}$$

Multiply 2 times 5.

$$p=\frac{48±52}{10}$$

Now solve the equation p=\frac{48±52}{10} when ± is plus. Add 48 to 52.

$$p=\frac{100}{10}$$

Divide 100 by 10.

$$p=10$$

Now solve the equation p=\frac{48±52}{10} when ± is minus. Subtract 52 from 48.

$$p=-\frac{4}{10}$$

Reduce the fraction \frac{-4}{10} to lowest terms by extracting and canceling out 2.

$$p=-\frac{2}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and -\frac{2}{5} for x_{2}.

$$5p^{2}-48p-20=5\left(p-10\right)\left(p-\left(-\frac{2}{5}\right)\right)$$

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

$$5p^{2}-48p-20=5\left(p-10\right)\left(p+\frac{2}{5}\right)$$

Add \frac{2}{5} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5p^{2}-48p-20=5\left(p-10\right)\times \frac{5p+2}{5}$$

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

$$5p^{2}-48p-20=\left(p-10\right)\left(5p+2\right)$$

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

$$x ^ 2 -\frac{48}{5}x -4 = 0$$

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

$$r + s = \frac{48}{5} rs = -4$$

Two numbers r and s sum up to \frac{48}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{48}{5} = \frac{24}{5}. 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>

$$r = \frac{24}{5} - u s = \frac{24}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(\frac{24}{5} - u) (\frac{24}{5} + u) = -4$$

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

$$\frac{576}{25} - u^2 = -4$$

Simplify the expression by subtracting \frac{576}{25} on both sides

$$-u^2 = -4-\frac{576}{25} = -\frac{676}{25}$$

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

$$u^2 = \frac{676}{25} u = \pm\sqrt{\frac{676}{25}} = \pm \frac{26}{5}$$

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

$$r =\frac{24}{5} - \frac{26}{5} = -0.400 s = \frac{24}{5} + \frac{26}{5} = 10$$

Evaluate

$$\left(p-10\right)\left(5p+2\right)$$