Solve 2b^2+15b+28

$$2 b ^ { 2 } + 15 b + 28$$

Factor

$$\left(b+4\right)\left(2b+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2b^{2}+pb+qb+28. To find p and q, set up a system to be solved.

$$p+q=15 pq=2\times 28=56$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 56.

$$1,56 2,28 4,14 7,8$$

Calculate the sum for each pair.

$$1+56=57 2+28=30 4+14=18 7+8=15$$

The solution is the pair that gives sum 15.

$$p=7 q=8$$

Rewrite 2b^{2}+15b+28 as \left(2b^{2}+7b\right)+\left(8b+28\right).

$$\left(2b^{2}+7b\right)+\left(8b+28\right)$$

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

$$b\left(2b+7\right)+4\left(2b+7\right)$$

Factor out common term 2b+7 by using distributive property.

$$\left(2b+7\right)\left(b+4\right)$$

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.

$$2b^{2}+15b+28=0$$

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.

$$b=\frac{-15±\sqrt{15^{2}-4\times 2\times 28}}{2\times 2}$$

Square 15.

$$b=\frac{-15±\sqrt{225-4\times 2\times 28}}{2\times 2}$$

Multiply -4 times 2.

$$b=\frac{-15±\sqrt{225-8\times 28}}{2\times 2}$$

Multiply -8 times 28.

$$b=\frac{-15±\sqrt{225-224}}{2\times 2}$$

Add 225 to -224.

$$b=\frac{-15±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$b=\frac{-15±1}{2\times 2}$$

Multiply 2 times 2.

$$b=\frac{-15±1}{4}$$

Now solve the equation b=\frac{-15±1}{4} when ± is plus. Add -15 to 1.

$$b=-\frac{14}{4}$$

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

$$b=-\frac{7}{2}$$

Now solve the equation b=\frac{-15±1}{4} when ± is minus. Subtract 1 from -15.

$$b=-\frac{16}{4}$$

Divide -16 by 4.

$$b=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{7}{2} for x_{1} and -4 for x_{2}.

$$2b^{2}+15b+28=2\left(b-\left(-\frac{7}{2}\right)\right)\left(b-\left(-4\right)\right)$$

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

$$2b^{2}+15b+28=2\left(b+\frac{7}{2}\right)\left(b+4\right)$$

Add \frac{7}{2} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$2b^{2}+15b+28=2\times \frac{2b+7}{2}\left(b+4\right)$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2b^{2}+15b+28=\left(2b+7\right)\left(b+4\right)$$

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 2

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

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

$$r + s = -\frac{15}{2} rs = 14$$

Two numbers r and s sum up to -\frac{15}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{15}{2} = -\frac{15}{4}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

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

To solve for unknown quantity u, substitute these in the product equation rs = 14

$$(-\frac{15}{4} - u) (-\frac{15}{4} + u) = 14$$

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

$$\frac{225}{16} - u^2 = 14$$

Simplify the expression by subtracting \frac{225}{16} on both sides

$$-u^2 = 14-\frac{225}{16} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =-\frac{15}{4} - \frac{1}{4} = -4 s = -\frac{15}{4} + \frac{1}{4} = -3.500$$

Evaluate

$$\left(b+4\right)\left(2b+7\right)$$