Solve 4b^2+4b-15

$$4 b ^ { 2 } + 4 b - 15$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4b^{2}+pb+qb-15. To find p and q, set up a system to be solved.

$$p+q=4 pq=4\left(-15\right)=-60$$

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 -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 4.

$$p=-6 q=10$$

Rewrite 4b^{2}+4b-15 as \left(4b^{2}-6b\right)+\left(10b-15\right).

$$\left(4b^{2}-6b\right)+\left(10b-15\right)$$

Factor out 2b in the first and 5 in the second group.

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

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

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

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.

$$4b^{2}+4b-15=0$$

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.

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

Square 4.

$$b=\frac{-4±\sqrt{16-4\times 4\left(-15\right)}}{2\times 4}$$

Multiply -4 times 4.

$$b=\frac{-4±\sqrt{16-16\left(-15\right)}}{2\times 4}$$

Multiply -16 times -15.

$$b=\frac{-4±\sqrt{16+240}}{2\times 4}$$

Add 16 to 240.

$$b=\frac{-4±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

$$b=\frac{-4±16}{2\times 4}$$

Multiply 2 times 4.

$$b=\frac{-4±16}{8}$$

Now solve the equation b=\frac{-4±16}{8} when ± is plus. Add -4 to 16.

$$b=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation b=\frac{-4±16}{8} when ± is minus. Subtract 16 from -4.

$$b=-\frac{20}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

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

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

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

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

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

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

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

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

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

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

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

Multiply 2 times 2.

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

Cancel out 4, the greatest common factor in 4 and 4.

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

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 4

$$x ^ 2 +1x -\frac{15}{4} = 0$$

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

$$r + s = -1 rs = -\frac{15}{4}$$

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 = -\frac{15}{4}

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -\frac{15}{4}$$

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

$$\frac{1}{4} - u^2 = -\frac{15}{4}$$

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

$$-u^2 = -\frac{15}{4}-\frac{1}{4} = -4$$

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

$$r =-\frac{1}{2} - 2 = -2.500 s = -\frac{1}{2} + 2 = 1.500$$

Evaluate

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