Solve -6b^2+b+12?

$$- 6 b ^ { 2 } + b + 12 ?$$

Factor

$$\left(-3b-4\right)\left(2b-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -6b^{2}+pb+qb+12. To find p and q, set up a system to be solved.

$$p+q=1 pq=-6\times 12=-72$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 1.

$$p=9 q=-8$$

Rewrite -6b^{2}+b+12 as \left(-6b^{2}+9b\right)+\left(-8b+12\right).

$$\left(-6b^{2}+9b\right)+\left(-8b+12\right)$$

Factor out -3b in the first and -4 in the second group.

$$-3b\left(2b-3\right)-4\left(2b-3\right)$$

Factor out common term 2b-3 by using distributive property.

$$\left(2b-3\right)\left(-3b-4\right)$$

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.

$$-6b^{2}+b+12=0$$

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.

$$b=\frac{-1±\sqrt{1^{2}-4\left(-6\right)\times 12}}{2\left(-6\right)}$$

Square 1.

$$b=\frac{-1±\sqrt{1-4\left(-6\right)\times 12}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$b=\frac{-1±\sqrt{1+24\times 12}}{2\left(-6\right)}$$

Multiply 24 times 12.

$$b=\frac{-1±\sqrt{1+288}}{2\left(-6\right)}$$

Add 1 to 288.

$$b=\frac{-1±\sqrt{289}}{2\left(-6\right)}$$

Take the square root of 289.

$$b=\frac{-1±17}{2\left(-6\right)}$$

Multiply 2 times -6.

$$b=\frac{-1±17}{-12}$$

Now solve the equation b=\frac{-1±17}{-12} when ± is plus. Add -1 to 17.

$$b=\frac{16}{-12}$$

Reduce the fraction \frac{16}{-12} to lowest terms by extracting and canceling out 4.

$$b=-\frac{4}{3}$$

Now solve the equation b=\frac{-1±17}{-12} when ± is minus. Subtract 17 from -1.

$$b=-\frac{18}{-12}$$

Reduce the fraction \frac{-18}{-12} to lowest terms by extracting and canceling out 6.

$$b=\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{4}{3} for x_{1} and \frac{3}{2} for x_{2}.

$$-6b^{2}+b+12=-6\left(b-\left(-\frac{4}{3}\right)\right)\left(b-\frac{3}{2}\right)$$

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

$$-6b^{2}+b+12=-6\left(b+\frac{4}{3}\right)\left(b-\frac{3}{2}\right)$$

Add \frac{4}{3} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-6b^{2}+b+12=-6\times \frac{-3b-4}{-3}\left(b-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from b by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6b^{2}+b+12=-6\times \frac{-3b-4}{-3}\times \frac{-2b+3}{-2}$$

Multiply \frac{-3b-4}{-3} times \frac{-2b+3}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-6b^{2}+b+12=-6\times \frac{\left(-3b-4\right)\left(-2b+3\right)}{-3\left(-2\right)}$$

Multiply -3 times -2.

$$-6b^{2}+b+12=-6\times \frac{\left(-3b-4\right)\left(-2b+3\right)}{6}$$

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

$$-6b^{2}+b+12=-\left(-3b-4\right)\left(-2b+3\right)$$

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.

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

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

$$r + s = \frac{1}{6} rs = -2$$

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 = -2

$$(\frac{1}{12} - u) (\frac{1}{12} + u) = -2$$

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

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

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

$$-u^2 = -2-\frac{1}{144} = -\frac{289}{144}$$

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

$$u^2 = \frac{289}{144} u = \pm\sqrt{\frac{289}{144}} = \pm \frac{17}{12}$$

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

$$r =\frac{1}{12} - \frac{17}{12} = -1.333 s = \frac{1}{12} + \frac{17}{12} = 1.500$$

Evaluate

$$12+b-6b^{2}$$