Solve 9p^2+59p+30

$$9 p ^ { 2 } + 59 p + 30$$

Factor

$$\left(p+6\right)\left(9p+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9p^{2}+ap+bp+30. To find a and b, set up a system to be solved.

$$a+b=59 ab=9\times 30=270$$

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 270.

$$1,270 2,135 3,90 5,54 6,45 9,30 10,27 15,18$$

Calculate the sum for each pair.

$$1+270=271 2+135=137 3+90=93 5+54=59 6+45=51 9+30=39 10+27=37 15+18=33$$

The solution is the pair that gives sum 59.

$$a=5 b=54$$

Rewrite 9p^{2}+59p+30 as \left(9p^{2}+5p\right)+\left(54p+30\right).

$$\left(9p^{2}+5p\right)+\left(54p+30\right)$$

Factor out p in the first and 6 in the second group.

$$p\left(9p+5\right)+6\left(9p+5\right)$$

Factor out common term 9p+5 by using distributive property.

$$\left(9p+5\right)\left(p+6\right)$$

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.

$$9p^{2}+59p+30=0$$

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.

$$p=\frac{-59±\sqrt{59^{2}-4\times 9\times 30}}{2\times 9}$$

Square 59.

$$p=\frac{-59±\sqrt{3481-4\times 9\times 30}}{2\times 9}$$

Multiply -4 times 9.

$$p=\frac{-59±\sqrt{3481-36\times 30}}{2\times 9}$$

Multiply -36 times 30.

$$p=\frac{-59±\sqrt{3481-1080}}{2\times 9}$$

Add 3481 to -1080.

$$p=\frac{-59±\sqrt{2401}}{2\times 9}$$

Take the square root of 2401.

$$p=\frac{-59±49}{2\times 9}$$

Multiply 2 times 9.

$$p=\frac{-59±49}{18}$$

Now solve the equation p=\frac{-59±49}{18} when ± is plus. Add -59 to 49.

$$p=-\frac{10}{18}$$

Reduce the fraction \frac{-10}{18} to lowest terms by extracting and canceling out 2.

$$p=-\frac{5}{9}$$

Now solve the equation p=\frac{-59±49}{18} when ± is minus. Subtract 49 from -59.

$$p=-\frac{108}{18}$$

Divide -108 by 18.

$$p=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{9} for x_{1} and -6 for x_{2}.

$$9p^{2}+59p+30=9\left(p-\left(-\frac{5}{9}\right)\right)\left(p-\left(-6\right)\right)$$

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

$$9p^{2}+59p+30=9\left(p+\frac{5}{9}\right)\left(p+6\right)$$

Add \frac{5}{9} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9p^{2}+59p+30=9\times \frac{9p+5}{9}\left(p+6\right)$$

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

$$9p^{2}+59p+30=\left(9p+5\right)\left(p+6\right)$$

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

$$x ^ 2 +\frac{59}{9}x +\frac{10}{3} = 0$$

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

$$r + s = -\frac{59}{9} rs = \frac{10}{3}$$

Two numbers r and s sum up to -\frac{59}{9} exactly when the average of the two numbers is \frac{1}{2}*-\frac{59}{9} = -\frac{59}{18}. 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>

$$r = -\frac{59}{18} - u s = -\frac{59}{18} + u$$

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

$$(-\frac{59}{18} - u) (-\frac{59}{18} + u) = \frac{10}{3}$$

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

$$\frac{3481}{324} - u^2 = \frac{10}{3}$$

Simplify the expression by subtracting \frac{3481}{324} on both sides

$$-u^2 = \frac{10}{3}-\frac{3481}{324} = -\frac{2401}{324}$$

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

$$u^2 = \frac{2401}{324} u = \pm\sqrt{\frac{2401}{324}} = \pm \frac{49}{18}$$

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

$$r =-\frac{59}{18} - \frac{49}{18} = -6 s = -\frac{59}{18} + \frac{49}{18} = -0.556$$

Evaluate

$$\left(p+6\right)\left(9p+5\right)$$