Solve -6P^2+P+1

$$- 6 P ^ { 2 } + P + 1$$

Factor

$$\left(-3P-1\right)\left(2P-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -6P^{2}+aP+bP+1. To find a and b, set up a system to be solved.

$$a+b=1 ab=-6=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=3 b=-2$$

Rewrite -6P^{2}+P+1 as \left(-6P^{2}+3P\right)+\left(-2P+1\right).

$$\left(-6P^{2}+3P\right)+\left(-2P+1\right)$$

Factor out -3P in the first and -1 in the second group.

$$-3P\left(2P-1\right)-\left(2P-1\right)$$

Factor out common term 2P-1 by using distributive property.

$$\left(2P-1\right)\left(-3P-1\right)$$

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.

$$-6P^{2}+P+1=0$$

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.

$$P=\frac{-1±\sqrt{1^{2}-4\left(-6\right)}}{2\left(-6\right)}$$

Square 1.

$$P=\frac{-1±\sqrt{1-4\left(-6\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$P=\frac{-1±\sqrt{1+24}}{2\left(-6\right)}$$

Add 1 to 24.

$$P=\frac{-1±\sqrt{25}}{2\left(-6\right)}$$

Take the square root of 25.

$$P=\frac{-1±5}{2\left(-6\right)}$$

Multiply 2 times -6.

$$P=\frac{-1±5}{-12}$$

Now solve the equation P=\frac{-1±5}{-12} when ± is plus. Add -1 to 5.

$$P=\frac{4}{-12}$$

Reduce the fraction \frac{4}{-12} to lowest terms by extracting and canceling out 4.

$$P=-\frac{1}{3}$$

Now solve the equation P=\frac{-1±5}{-12} when ± is minus. Subtract 5 from -1.

$$P=-\frac{6}{-12}$$

Reduce the fraction \frac{-6}{-12} to lowest terms by extracting and canceling out 6.

$$P=\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{3} for x_{1} and \frac{1}{2} for x_{2}.

$$-6P^{2}+P+1=-6\left(P-\left(-\frac{1}{3}\right)\right)\left(P-\frac{1}{2}\right)$$

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

$$-6P^{2}+P+1=-6\left(P+\frac{1}{3}\right)\left(P-\frac{1}{2}\right)$$

Add \frac{1}{3} to P by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-6P^{2}+P+1=-6\times \frac{-3P-1}{-3}\left(P-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from P by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6P^{2}+P+1=-6\times \frac{-3P-1}{-3}\times \frac{-2P+1}{-2}$$

Multiply \frac{-3P-1}{-3} times \frac{-2P+1}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-6P^{2}+P+1=-6\times \frac{\left(-3P-1\right)\left(-2P+1\right)}{-3\left(-2\right)}$$

Multiply -3 times -2.

$$-6P^{2}+P+1=-6\times \frac{\left(-3P-1\right)\left(-2P+1\right)}{6}$$

Cancel out 6, the greatest common factor in -6 and 6.

$$-6P^{2}+P+1=-\left(-3P-1\right)\left(-2P+1\right)$$

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.

$$x ^ 2 -\frac{1}{6}x -\frac{1}{6} = 0$$

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

$$r + s = \frac{1}{6} rs = -\frac{1}{6}$$

Two numbers r and s sum up to \frac{1}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{6} = \frac{1}{12}. 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>

$$r = \frac{1}{12} - u s = \frac{1}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{6}

$$(\frac{1}{12} - u) (\frac{1}{12} + u) = -\frac{1}{6}$$

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

$$\frac{1}{144} - u^2 = -\frac{1}{6}$$

Simplify the expression by subtracting \frac{1}{144} on both sides

$$-u^2 = -\frac{1}{6}-\frac{1}{144} = -\frac{25}{144}$$

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

$$u^2 = \frac{25}{144} u = \pm\sqrt{\frac{25}{144}} = \pm \frac{5}{12}$$

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

$$r =\frac{1}{12} - \frac{5}{12} = -0.333 s = \frac{1}{12} + \frac{5}{12} = 0.500$$

Evaluate

$$1+P-6P^{2}$$