Solve 6a^2+22a+20

$$6 a ^ { 2 } + 22 a + 20$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(3a^{2}+11a+10\right)$$

Consider 3a^{2}+11a+10. Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa+10. To find p and q, set up a system to be solved.

$$p+q=11 pq=3\times 10=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 11.

$$p=5 q=6$$

Rewrite 3a^{2}+11a+10 as \left(3a^{2}+5a\right)+\left(6a+10\right).

$$\left(3a^{2}+5a\right)+\left(6a+10\right)$$

Factor out a in the first and 2 in the second group.

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

Factor out common term 3a+5 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$6a^{2}+22a+20=0$$

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.

$$a=\frac{-22±\sqrt{22^{2}-4\times 6\times 20}}{2\times 6}$$

Square 22.

$$a=\frac{-22±\sqrt{484-4\times 6\times 20}}{2\times 6}$$

Multiply -4 times 6.

$$a=\frac{-22±\sqrt{484-24\times 20}}{2\times 6}$$

Multiply -24 times 20.

$$a=\frac{-22±\sqrt{484-480}}{2\times 6}$$

Add 484 to -480.

$$a=\frac{-22±\sqrt{4}}{2\times 6}$$

Take the square root of 4.

$$a=\frac{-22±2}{2\times 6}$$

Multiply 2 times 6.

$$a=\frac{-22±2}{12}$$

Now solve the equation a=\frac{-22±2}{12} when ± is plus. Add -22 to 2.

$$a=-\frac{20}{12}$$

Reduce the fraction \frac{-20}{12} to lowest terms by extracting and canceling out 4.

$$a=-\frac{5}{3}$$

Now solve the equation a=\frac{-22±2}{12} when ± is minus. Subtract 2 from -22.

$$a=-\frac{24}{12}$$

Divide -24 by 12.

$$a=-2$$

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

$$6a^{2}+22a+20=6\left(a-\left(-\frac{5}{3}\right)\right)\left(a-\left(-2\right)\right)$$

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

$$6a^{2}+22a+20=6\left(a+\frac{5}{3}\right)\left(a+2\right)$$

Add \frac{5}{3} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6a^{2}+22a+20=6\times \frac{3a+5}{3}\left(a+2\right)$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6a^{2}+22a+20=2\left(3a+5\right)\left(a+2\right)$$

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 6

$$x ^ 2 +\frac{11}{3}x +\frac{10}{3} = 0$$

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

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

Two numbers r and s sum up to -\frac{11}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{3} = -\frac{11}{6}. 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>

$$r = -\frac{11}{6} - u s = -\frac{11}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{10}{3}

$$(-\frac{11}{6} - u) (-\frac{11}{6} + u) = \frac{10}{3}$$

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

$$\frac{121}{36} - u^2 = \frac{10}{3}$$

Simplify the expression by subtracting \frac{121}{36} on both sides

$$-u^2 = \frac{10}{3}-\frac{121}{36} = -\frac{1}{36}$$

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

$$u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}$$

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

$$r =-\frac{11}{6} - \frac{1}{6} = -2.000 s = -\frac{11}{6} + \frac{1}{6} = -1.667$$

Evaluate

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