Solve 15b^2+14b-8

$$15 b ^ { 2 } + 14 b - 8$$

Factor

$$\left(5b-2\right)\left(3b+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15b^{2}+pb+qb-8. To find p and q, set up a system to be solved.

$$p+q=14 pq=15\left(-8\right)=-120$$

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 -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 14.

$$p=-6 q=20$$

Rewrite 15b^{2}+14b-8 as \left(15b^{2}-6b\right)+\left(20b-8\right).

$$\left(15b^{2}-6b\right)+\left(20b-8\right)$$

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

$$3b\left(5b-2\right)+4\left(5b-2\right)$$

Factor out common term 5b-2 by using distributive property.

$$\left(5b-2\right)\left(3b+4\right)$$

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.

$$15b^{2}+14b-8=0$$

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.

$$b=\frac{-14±\sqrt{14^{2}-4\times 15\left(-8\right)}}{2\times 15}$$

Square 14.

$$b=\frac{-14±\sqrt{196-4\times 15\left(-8\right)}}{2\times 15}$$

Multiply -4 times 15.

$$b=\frac{-14±\sqrt{196-60\left(-8\right)}}{2\times 15}$$

Multiply -60 times -8.

$$b=\frac{-14±\sqrt{196+480}}{2\times 15}$$

Add 196 to 480.

$$b=\frac{-14±\sqrt{676}}{2\times 15}$$

Take the square root of 676.

$$b=\frac{-14±26}{2\times 15}$$

Multiply 2 times 15.

$$b=\frac{-14±26}{30}$$

Now solve the equation b=\frac{-14±26}{30} when ± is plus. Add -14 to 26.

$$b=\frac{12}{30}$$

Reduce the fraction \frac{12}{30} to lowest terms by extracting and canceling out 6.

$$b=\frac{2}{5}$$

Now solve the equation b=\frac{-14±26}{30} when ± is minus. Subtract 26 from -14.

$$b=-\frac{40}{30}$$

Reduce the fraction \frac{-40}{30} to lowest terms by extracting and canceling out 10.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{5} for x_{1} and -\frac{4}{3} for x_{2}.

$$15b^{2}+14b-8=15\left(b-\frac{2}{5}\right)\left(b-\left(-\frac{4}{3}\right)\right)$$

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

$$15b^{2}+14b-8=15\left(b-\frac{2}{5}\right)\left(b+\frac{4}{3}\right)$$

Subtract \frac{2}{5} from b by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15b^{2}+14b-8=15\times \frac{5b-2}{5}\left(b+\frac{4}{3}\right)$$

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

$$15b^{2}+14b-8=15\times \frac{5b-2}{5}\times \frac{3b+4}{3}$$

Multiply \frac{5b-2}{5} times \frac{3b+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15b^{2}+14b-8=15\times \frac{\left(5b-2\right)\left(3b+4\right)}{5\times 3}$$

Multiply 5 times 3.

$$15b^{2}+14b-8=15\times \frac{\left(5b-2\right)\left(3b+4\right)}{15}$$

Cancel out 15, the greatest common factor in 15 and 15.

$$15b^{2}+14b-8=\left(5b-2\right)\left(3b+4\right)$$

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 15

$$x ^ 2 +\frac{14}{15}x -\frac{8}{15} = 0$$

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

$$r + s = -\frac{14}{15} rs = -\frac{8}{15}$$

Two numbers r and s sum up to -\frac{14}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{14}{15} = -\frac{7}{15}. 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>

$$r = -\frac{7}{15} - u s = -\frac{7}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{8}{15}

$$(-\frac{7}{15} - u) (-\frac{7}{15} + u) = -\frac{8}{15}$$

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

$$\frac{49}{225} - u^2 = -\frac{8}{15}$$

Simplify the expression by subtracting \frac{49}{225} on both sides

$$-u^2 = -\frac{8}{15}-\frac{49}{225} = -\frac{169}{225}$$

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

$$u^2 = \frac{169}{225} u = \pm\sqrt{\frac{169}{225}} = \pm \frac{13}{15}$$

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

$$r =-\frac{7}{15} - \frac{13}{15} = -1.333 s = -\frac{7}{15} + \frac{13}{15} = 0.400$$

Evaluate

$$\left(5b-2\right)\left(3b+4\right)$$