Solve 4a^2+15a+9=

$$4 a ^ { 2 } + 15 a + 9 =$$

Factor

$$\left(a+3\right)\left(4a+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4a^{2}+pa+qa+9. To find p and q, set up a system to be solved.

$$p+q=15 pq=4\times 9=36$$

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 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 15.

$$p=3 q=12$$

Rewrite 4a^{2}+15a+9 as \left(4a^{2}+3a\right)+\left(12a+9\right).

$$\left(4a^{2}+3a\right)+\left(12a+9\right)$$

Factor out a in the first and 3 in the second group.

$$a\left(4a+3\right)+3\left(4a+3\right)$$

Factor out common term 4a+3 by using distributive property.

$$\left(4a+3\right)\left(a+3\right)$$

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.

$$4a^{2}+15a+9=0$$

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.

$$a=\frac{-15±\sqrt{15^{2}-4\times 4\times 9}}{2\times 4}$$

Square 15.

$$a=\frac{-15±\sqrt{225-4\times 4\times 9}}{2\times 4}$$

Multiply -4 times 4.

$$a=\frac{-15±\sqrt{225-16\times 9}}{2\times 4}$$

Multiply -16 times 9.

$$a=\frac{-15±\sqrt{225-144}}{2\times 4}$$

Add 225 to -144.

$$a=\frac{-15±\sqrt{81}}{2\times 4}$$

Take the square root of 81.

$$a=\frac{-15±9}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{-15±9}{8}$$

Now solve the equation a=\frac{-15±9}{8} when ± is plus. Add -15 to 9.

$$a=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

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

Now solve the equation a=\frac{-15±9}{8} when ± is minus. Subtract 9 from -15.

$$a=-\frac{24}{8}$$

Divide -24 by 8.

$$a=-3$$

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

$$4a^{2}+15a+9=4\left(a-\left(-\frac{3}{4}\right)\right)\left(a-\left(-3\right)\right)$$

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

$$4a^{2}+15a+9=4\left(a+\frac{3}{4}\right)\left(a+3\right)$$

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

$$4a^{2}+15a+9=4\times \frac{4a+3}{4}\left(a+3\right)$$

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

$$4a^{2}+15a+9=\left(4a+3\right)\left(a+3\right)$$

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

$$x ^ 2 +\frac{15}{4}x +\frac{9}{4} = 0$$

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

$$r + s = -\frac{15}{4} rs = \frac{9}{4}$$

Two numbers r and s sum up to -\frac{15}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{15}{4} = -\frac{15}{8}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{4}

$$(-\frac{15}{8} - u) (-\frac{15}{8} + u) = \frac{9}{4}$$

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

$$\frac{225}{64} - u^2 = \frac{9}{4}$$

Simplify the expression by subtracting \frac{225}{64} on both sides

$$-u^2 = \frac{9}{4}-\frac{225}{64} = -\frac{81}{64}$$

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

$$u^2 = \frac{81}{64} u = \pm\sqrt{\frac{81}{64}} = \pm \frac{9}{8}$$

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

$$r =-\frac{15}{8} - \frac{9}{8} = -3 s = -\frac{15}{8} + \frac{9}{8} = -0.750$$

Evaluate

$$\left(a+3\right)\left(4a+3\right)$$