Solve 9x^2+39x+42=

$$9 x ^ { 2 } + 39 x + 42 =$$

Factor

$$3\left(x+2\right)\left(3x+7\right)$$

Solution Steps

Factor out 3.

$$3\left(3x^{2}+13x+14\right)$$

Consider 3x^{2}+13x+14. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=13 ab=3\times 14=42$$

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 42.

$$1,42 2,21 3,14 6,7$$

Calculate the sum for each pair.

$$1+42=43 2+21=23 3+14=17 6+7=13$$

The solution is the pair that gives sum 13.

$$a=6 b=7$$

Rewrite 3x^{2}+13x+14 as \left(3x^{2}+6x\right)+\left(7x+14\right).

$$\left(3x^{2}+6x\right)+\left(7x+14\right)$$

Factor out 3x in the first and 7 in the second group.

$$3x\left(x+2\right)+7\left(x+2\right)$$

Factor out common term x+2 by using distributive property.

$$\left(x+2\right)\left(3x+7\right)$$

Rewrite the complete factored expression.

$$3\left(x+2\right)\left(3x+7\right)$$

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.

$$9x^{2}+39x+42=0$$

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.

$$x=\frac{-39±\sqrt{39^{2}-4\times 9\times 42}}{2\times 9}$$

Square 39.

$$x=\frac{-39±\sqrt{1521-4\times 9\times 42}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-39±\sqrt{1521-36\times 42}}{2\times 9}$$

Multiply -36 times 42.

$$x=\frac{-39±\sqrt{1521-1512}}{2\times 9}$$

Add 1521 to -1512.

$$x=\frac{-39±\sqrt{9}}{2\times 9}$$

Take the square root of 9.

$$x=\frac{-39±3}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-39±3}{18}$$

Now solve the equation x=\frac{-39±3}{18} when ± is plus. Add -39 to 3.

$$x=-\frac{36}{18}$$

Divide -36 by 18.

$$x=-2$$

Now solve the equation x=\frac{-39±3}{18} when ± is minus. Subtract 3 from -39.

$$x=-\frac{42}{18}$$

Reduce the fraction \frac{-42}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{7}{3}$$

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

$$9x^{2}+39x+42=9\left(x-\left(-2\right)\right)\left(x-\left(-\frac{7}{3}\right)\right)$$

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

$$9x^{2}+39x+42=9\left(x+2\right)\left(x+\frac{7}{3}\right)$$

Add \frac{7}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}+39x+42=9\left(x+2\right)\times \frac{3x+7}{3}$$

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

$$9x^{2}+39x+42=3\left(x+2\right)\left(3x+7\right)$$

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

$$x ^ 2 +\frac{13}{3}x +\frac{14}{3} = 0$$

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

$$r + s = -\frac{13}{3} rs = \frac{14}{3}$$

Two numbers r and s sum up to -\frac{13}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{3} = -\frac{13}{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>

$$r = -\frac{13}{6} - u s = -\frac{13}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{14}{3}

$$(-\frac{13}{6} - u) (-\frac{13}{6} + u) = \frac{14}{3}$$

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

$$\frac{169}{36} - u^2 = \frac{14}{3}$$

Simplify the expression by subtracting \frac{169}{36} on both sides

$$-u^2 = \frac{14}{3}-\frac{169}{36} = -\frac{1}{36}$$

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.

$$r =-\frac{13}{6} - \frac{1}{6} = -2.333 s = -\frac{13}{6} + \frac{1}{6} = -2.000$$

Evaluate

$$3\left(x+2\right)\left(3x+7\right)$$