Solve 25p^2-5p-2

$$25 p ^ { 2 } - 5 p - 2$$

Factor

$$\left(5p-2\right)\left(5p+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 25p^{2}+ap+bp-2. To find a and b, set up a system to be solved.

$$a+b=-5 ab=25\left(-2\right)=-50$$

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 -50.

$$1,-50 2,-25 5,-10$$

Calculate the sum for each pair.

$$1-50=-49 2-25=-23 5-10=-5$$

The solution is the pair that gives sum -5.

$$a=-10 b=5$$

Rewrite 25p^{2}-5p-2 as \left(25p^{2}-10p\right)+\left(5p-2\right).

$$\left(25p^{2}-10p\right)+\left(5p-2\right)$$

Factor out 5p in 25p^{2}-10p.

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

Factor out common term 5p-2 by using distributive property.

$$\left(5p-2\right)\left(5p+1\right)$$

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.

$$25p^{2}-5p-2=0$$

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.

$$p=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 25\left(-2\right)}}{2\times 25}$$

Square -5.

$$p=\frac{-\left(-5\right)±\sqrt{25-4\times 25\left(-2\right)}}{2\times 25}$$

Multiply -4 times 25.

$$p=\frac{-\left(-5\right)±\sqrt{25-100\left(-2\right)}}{2\times 25}$$

Multiply -100 times -2.

$$p=\frac{-\left(-5\right)±\sqrt{25+200}}{2\times 25}$$

Add 25 to 200.

$$p=\frac{-\left(-5\right)±\sqrt{225}}{2\times 25}$$

Take the square root of 225.

$$p=\frac{-\left(-5\right)±15}{2\times 25}$$

The opposite of -5 is 5.

$$p=\frac{5±15}{2\times 25}$$

Multiply 2 times 25.

$$p=\frac{5±15}{50}$$

Now solve the equation p=\frac{5±15}{50} when ± is plus. Add 5 to 15.

$$p=\frac{20}{50}$$

Reduce the fraction \frac{20}{50} to lowest terms by extracting and canceling out 10.

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

Now solve the equation p=\frac{5±15}{50} when ± is minus. Subtract 15 from 5.

$$p=-\frac{10}{50}$$

Reduce the fraction \frac{-10}{50} to lowest terms by extracting and canceling out 10.

$$p=-\frac{1}{5}$$

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

$$25p^{2}-5p-2=25\left(p-\frac{2}{5}\right)\left(p-\left(-\frac{1}{5}\right)\right)$$

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

$$25p^{2}-5p-2=25\left(p-\frac{2}{5}\right)\left(p+\frac{1}{5}\right)$$

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

$$25p^{2}-5p-2=25\times \frac{5p-2}{5}\left(p+\frac{1}{5}\right)$$

Add \frac{1}{5} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25p^{2}-5p-2=25\times \frac{5p-2}{5}\times \frac{5p+1}{5}$$

Multiply \frac{5p-2}{5} times \frac{5p+1}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$25p^{2}-5p-2=25\times \frac{\left(5p-2\right)\left(5p+1\right)}{5\times 5}$$

Multiply 5 times 5.

$$25p^{2}-5p-2=25\times \frac{\left(5p-2\right)\left(5p+1\right)}{25}$$

Cancel out 25, the greatest common factor in 25 and 25.

$$25p^{2}-5p-2=\left(5p-2\right)\left(5p+1\right)$$

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 25

$$x ^ 2 -\frac{1}{5}x -\frac{2}{25} = 0$$

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

$$r + s = \frac{1}{5} rs = -\frac{2}{25}$$

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

$$r = \frac{1}{10} - u s = \frac{1}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{25}

$$(\frac{1}{10} - u) (\frac{1}{10} + u) = -\frac{2}{25}$$

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

$$\frac{1}{100} - u^2 = -\frac{2}{25}$$

Simplify the expression by subtracting \frac{1}{100} on both sides

$$-u^2 = -\frac{2}{25}-\frac{1}{100} = -\frac{9}{100}$$

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

$$u^2 = \frac{9}{100} u = \pm\sqrt{\frac{9}{100}} = \pm \frac{3}{10}$$

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

$$r =\frac{1}{10} - \frac{3}{10} = -0.200 s = \frac{1}{10} + \frac{3}{10} = 0.400$$

Evaluate

$$\left(5p-2\right)\left(5p+1\right)$$