Solve 9a^2+66a+21

$$9 a ^ { 2 } + 66 a + 21$$

Factor

$$3\left(a+7\right)\left(3a+1\right)$$

Solution Steps

Factor out 3.

$$3\left(3a^{2}+22a+7\right)$$

Consider 3a^{2}+22a+7. Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa+7. To find p and q, set up a system to be solved.

$$p+q=22 pq=3\times 7=21$$

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 21.

$$1,21 3,7$$

Calculate the sum for each pair.

$$1+21=22 3+7=10$$

The solution is the pair that gives sum 22.

$$p=1 q=21$$

Rewrite 3a^{2}+22a+7 as \left(3a^{2}+a\right)+\left(21a+7\right).

$$\left(3a^{2}+a\right)+\left(21a+7\right)$$

Factor out a in the first and 7 in the second group.

$$a\left(3a+1\right)+7\left(3a+1\right)$$

Factor out common term 3a+1 by using distributive property.

$$\left(3a+1\right)\left(a+7\right)$$

Rewrite the complete factored expression.

$$3\left(3a+1\right)\left(a+7\right)$$

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.

$$9a^{2}+66a+21=0$$

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.

$$a=\frac{-66±\sqrt{66^{2}-4\times 9\times 21}}{2\times 9}$$

Square 66.

$$a=\frac{-66±\sqrt{4356-4\times 9\times 21}}{2\times 9}$$

Multiply -4 times 9.

$$a=\frac{-66±\sqrt{4356-36\times 21}}{2\times 9}$$

Multiply -36 times 21.

$$a=\frac{-66±\sqrt{4356-756}}{2\times 9}$$

Add 4356 to -756.

$$a=\frac{-66±\sqrt{3600}}{2\times 9}$$

Take the square root of 3600.

$$a=\frac{-66±60}{2\times 9}$$

Multiply 2 times 9.

$$a=\frac{-66±60}{18}$$

Now solve the equation a=\frac{-66±60}{18} when ± is plus. Add -66 to 60.

$$a=-\frac{6}{18}$$

Reduce the fraction \frac{-6}{18} to lowest terms by extracting and canceling out 6.

$$a=-\frac{1}{3}$$

Now solve the equation a=\frac{-66±60}{18} when ± is minus. Subtract 60 from -66.

$$a=-\frac{126}{18}$$

Divide -126 by 18.

$$a=-7$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{3} for x_{1} and -7 for x_{2}.

$$9a^{2}+66a+21=9\left(a-\left(-\frac{1}{3}\right)\right)\left(a-\left(-7\right)\right)$$

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

$$9a^{2}+66a+21=9\left(a+\frac{1}{3}\right)\left(a+7\right)$$

Add \frac{1}{3} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9a^{2}+66a+21=9\times \frac{3a+1}{3}\left(a+7\right)$$

Cancel out 3, the greatest common factor in 9 and 3.

$$9a^{2}+66a+21=3\left(3a+1\right)\left(a+7\right)$$

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 9

$$x ^ 2 +\frac{22}{3}x +\frac{7}{3} = 0$$

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

$$r + s = -\frac{22}{3} rs = \frac{7}{3}$$

Two numbers r and s sum up to -\frac{22}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{22}{3} = -\frac{11}{3}. 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>

$$r = -\frac{11}{3} - u s = -\frac{11}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{3}

$$(-\frac{11}{3} - u) (-\frac{11}{3} + u) = \frac{7}{3}$$

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

$$\frac{121}{9} - u^2 = \frac{7}{3}$$

Simplify the expression by subtracting \frac{121}{9} on both sides

$$-u^2 = \frac{7}{3}-\frac{121}{9} = -\frac{100}{9}$$

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

$$u^2 = \frac{100}{9} u = \pm\sqrt{\frac{100}{9}} = \pm \frac{10}{3}$$

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

$$r =-\frac{11}{3} - \frac{10}{3} = -7 s = -\frac{11}{3} + \frac{10}{3} = -0.333$$

Evaluate

$$3\left(a+7\right)\left(3a+1\right)$$