Solve 2p^2+46p-100

$$2 p ^ { 2 } + 46 p - 100$$

Factor

$$2\left(p-2\right)\left(p+25\right)$$

Solution Steps

Factor out 2.

$$2\left(p^{2}+23p-50\right)$$

Consider p^{2}+23p-50. Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp-50. To find a and b, set up a system to be solved.

$$a+b=23 ab=1\left(-50\right)=-50$$

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 -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 23.

$$a=-2 b=25$$

Rewrite p^{2}+23p-50 as \left(p^{2}-2p\right)+\left(25p-50\right).

$$\left(p^{2}-2p\right)+\left(25p-50\right)$$

Factor out p in the first and 25 in the second group.

$$p\left(p-2\right)+25\left(p-2\right)$$

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

$$\left(p-2\right)\left(p+25\right)$$

Rewrite the complete factored expression.

$$2\left(p-2\right)\left(p+25\right)$$

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.

$$2p^{2}+46p-100=0$$

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.

$$p=\frac{-46±\sqrt{46^{2}-4\times 2\left(-100\right)}}{2\times 2}$$

Square 46.

$$p=\frac{-46±\sqrt{2116-4\times 2\left(-100\right)}}{2\times 2}$$

Multiply -4 times 2.

$$p=\frac{-46±\sqrt{2116-8\left(-100\right)}}{2\times 2}$$

Multiply -8 times -100.

$$p=\frac{-46±\sqrt{2116+800}}{2\times 2}$$

Add 2116 to 800.

$$p=\frac{-46±\sqrt{2916}}{2\times 2}$$

Take the square root of 2916.

$$p=\frac{-46±54}{2\times 2}$$

Multiply 2 times 2.

$$p=\frac{-46±54}{4}$$

Now solve the equation p=\frac{-46±54}{4} when ± is plus. Add -46 to 54.

$$p=\frac{8}{4}$$

Divide 8 by 4.

$$p=2$$

Now solve the equation p=\frac{-46±54}{4} when ± is minus. Subtract 54 from -46.

$$p=-\frac{100}{4}$$

Divide -100 by 4.

$$p=-25$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -25 for x_{2}.

$$2p^{2}+46p-100=2\left(p-2\right)\left(p-\left(-25\right)\right)$$

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

$$2p^{2}+46p-100=2\left(p-2\right)\left(p+25\right)$$

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 2

$$x ^ 2 +23x -50 = 0$$

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

$$r + s = -23 rs = -50$$

Two numbers r and s sum up to -23 exactly when the average of the two numbers is \frac{1}{2}*-23 = -\frac{23}{2}. 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>

$$r = -\frac{23}{2} - u s = -\frac{23}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -50

$$(-\frac{23}{2} - u) (-\frac{23}{2} + u) = -50$$

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

$$\frac{529}{4} - u^2 = -50$$

Simplify the expression by subtracting \frac{529}{4} on both sides

$$-u^2 = -50-\frac{529}{4} = -\frac{729}{4}$$

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

$$u^2 = \frac{729}{4} u = \pm\sqrt{\frac{729}{4}} = \pm \frac{27}{2}$$

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

$$r =-\frac{23}{2} - \frac{27}{2} = -25 s = -\frac{23}{2} + \frac{27}{2} = 2$$

Evaluate

$$2\left(p-2\right)\left(p+25\right)$$