Solve 3b^2-10b-8

$$3 b ^ { 2 } - 10 b - 8$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3b^{2}+pb+qb-8. To find p and q, set up a system to be solved.

$$p+q=-10 pq=3\left(-8\right)=-24$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -10.

$$p=-12 q=2$$

Rewrite 3b^{2}-10b-8 as \left(3b^{2}-12b\right)+\left(2b-8\right).

$$\left(3b^{2}-12b\right)+\left(2b-8\right)$$

Factor out 3b in the first and 2 in the second group.

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

Factor out common term b-4 by using distributive property.

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

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.

$$3b^{2}-10b-8=0$$

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.

$$b=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\left(-8\right)}}{2\times 3}$$

Square -10.

$$b=\frac{-\left(-10\right)±\sqrt{100-4\times 3\left(-8\right)}}{2\times 3}$$

Multiply -4 times 3.

$$b=\frac{-\left(-10\right)±\sqrt{100-12\left(-8\right)}}{2\times 3}$$

Multiply -12 times -8.

$$b=\frac{-\left(-10\right)±\sqrt{100+96}}{2\times 3}$$

Add 100 to 96.

$$b=\frac{-\left(-10\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$b=\frac{-\left(-10\right)±14}{2\times 3}$$

The opposite of -10 is 10.

$$b=\frac{10±14}{2\times 3}$$

Multiply 2 times 3.

$$b=\frac{10±14}{6}$$

Now solve the equation b=\frac{10±14}{6} when ± is plus. Add 10 to 14.

$$b=\frac{24}{6}$$

Divide 24 by 6.

$$b=4$$

Now solve the equation b=\frac{10±14}{6} when ± is minus. Subtract 14 from 10.

$$b=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

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

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

$$3b^{2}-10b-8=3\left(b-4\right)\left(b-\left(-\frac{2}{3}\right)\right)$$

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

$$3b^{2}-10b-8=3\left(b-4\right)\left(b+\frac{2}{3}\right)$$

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

$$3b^{2}-10b-8=3\left(b-4\right)\times \frac{3b+2}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3b^{2}-10b-8=\left(b-4\right)\left(3b+2\right)$$

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 3

$$x ^ 2 -\frac{10}{3}x -\frac{8}{3} = 0$$

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

$$r + s = \frac{10}{3} rs = -\frac{8}{3}$$

Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{3}. 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>

$$r = \frac{5}{3} - u s = \frac{5}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{8}{3}

$$(\frac{5}{3} - u) (\frac{5}{3} + u) = -\frac{8}{3}$$

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

$$\frac{25}{9} - u^2 = -\frac{8}{3}$$

Simplify the expression by subtracting \frac{25}{9} on both sides

$$-u^2 = -\frac{8}{3}-\frac{25}{9} = -\frac{49}{9}$$

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

$$u^2 = \frac{49}{9} u = \pm\sqrt{\frac{49}{9}} = \pm \frac{7}{3}$$

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

$$r =\frac{5}{3} - \frac{7}{3} = -0.667 s = \frac{5}{3} + \frac{7}{3} = 4$$

Evaluate

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