Solve 4y^2+16y+15

$$4 y ^ { 2 } + 16 y + 15$$

Factor

$$\left(2y+3\right)\left(2y+5\right)$$

Solution Steps

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

$$a+b=16 ab=4\times 15=60$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 16.

$$a=6 b=10$$

Rewrite 4y^{2}+16y+15 as \left(4y^{2}+6y\right)+\left(10y+15\right).

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

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

$$2y\left(2y+3\right)+5\left(2y+3\right)$$

Factor out common term 2y+3 by using distributive property.

$$\left(2y+3\right)\left(2y+5\right)$$

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.

$$4y^{2}+16y+15=0$$

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.

$$y=\frac{-16±\sqrt{16^{2}-4\times 4\times 15}}{2\times 4}$$

Square 16.

$$y=\frac{-16±\sqrt{256-4\times 4\times 15}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-16±\sqrt{256-16\times 15}}{2\times 4}$$

Multiply -16 times 15.

$$y=\frac{-16±\sqrt{256-240}}{2\times 4}$$

Add 256 to -240.

$$y=\frac{-16±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$y=\frac{-16±4}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{-16±4}{8}$$

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

$$y=-\frac{12}{8}$$

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

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

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

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

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

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

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}.

$$4y^{2}+16y+15=4\left(y-\left(-\frac{3}{2}\right)\right)\left(y-\left(-\frac{5}{2}\right)\right)$$

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

$$4y^{2}+16y+15=4\left(y+\frac{3}{2}\right)\left(y+\frac{5}{2}\right)$$

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

$$4y^{2}+16y+15=4\times \frac{2y+3}{2}\left(y+\frac{5}{2}\right)$$

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

$$4y^{2}+16y+15=4\times \frac{2y+3}{2}\times \frac{2y+5}{2}$$

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

$$4y^{2}+16y+15=4\times \frac{\left(2y+3\right)\left(2y+5\right)}{2\times 2}$$

Multiply 2 times 2.

$$4y^{2}+16y+15=4\times \frac{\left(2y+3\right)\left(2y+5\right)}{4}$$

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

$$4y^{2}+16y+15=\left(2y+3\right)\left(2y+5\right)$$

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

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

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

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

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -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>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{15}{4}

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

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

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

Simplify the expression by subtracting 4 on both sides

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

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

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

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

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

Evaluate

$$\left(2y+3\right)\left(2y+5\right)$$