Solve 2P^2+2P-4

$$2 P ^ { 2 } + 2 P - 4$$

Factor

$$2\left(P-1\right)\left(P+2\right)$$

Solution Steps

Factor out 2.

$$2\left(P^{2}+P-2\right)$$

Consider P^{2}+P-2. Factor the expression by grouping. First, the expression needs to be rewritten as P^{2}+aP+bP-2. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-2\right)=-2$$

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. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite P^{2}+P-2 as \left(P^{2}-P\right)+\left(2P-2\right).

$$\left(P^{2}-P\right)+\left(2P-2\right)$$

Factor out P in the first and 2 in the second group.

$$P\left(P-1\right)+2\left(P-1\right)$$

Factor out common term P-1 by using distributive property.

$$\left(P-1\right)\left(P+2\right)$$

Rewrite the complete factored expression.

$$2\left(P-1\right)\left(P+2\right)$$

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.

$$2P^{2}+2P-4=0$$

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.

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

Square 2.

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

Multiply -4 times 2.

$$P=\frac{-2±\sqrt{4-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$P=\frac{-2±\sqrt{4+32}}{2\times 2}$$

Add 4 to 32.

$$P=\frac{-2±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$P=\frac{-2±6}{2\times 2}$$

Multiply 2 times 2.

$$P=\frac{-2±6}{4}$$

Now solve the equation P=\frac{-2±6}{4} when ± is plus. Add -2 to 6.

$$P=\frac{4}{4}$$

Divide 4 by 4.

$$P=1$$

Now solve the equation P=\frac{-2±6}{4} when ± is minus. Subtract 6 from -2.

$$P=-\frac{8}{4}$$

Divide -8 by 4.

$$P=-2$$

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

$$2P^{2}+2P-4=2\left(P-1\right)\left(P-\left(-2\right)\right)$$

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

$$2P^{2}+2P-4=2\left(P-1\right)\left(P+2\right)$$

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 2

$$x ^ 2 +1x -2 = 0$$

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

$$r + s = -1 rs = -2$$

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

$$r = -\frac{1}{2} - u s = -\frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -2

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

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

$$\frac{1}{4} - u^2 = -2$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -2-\frac{1}{4} = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =-\frac{1}{2} - \frac{3}{2} = -2 s = -\frac{1}{2} + \frac{3}{2} = 1$$

Evaluate

$$2\left(P-1\right)\left(P+2\right)$$