Solve 6p^2+25p-9

$$6 p ^ { 2 } + 25 p - 9$$

Factor

$$\left(3p-1\right)\left(2p+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6p^{2}+ap+bp-9. To find a and b, set up a system to be solved.

$$a+b=25 ab=6\left(-9\right)=-54$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -54.

$$-1,54 -2,27 -3,18 -6,9$$

Calculate the sum for each pair.

$$-1+54=53 -2+27=25 -3+18=15 -6+9=3$$

The solution is the pair that gives sum 25.

$$a=-2 b=27$$

Rewrite 6p^{2}+25p-9 as \left(6p^{2}-2p\right)+\left(27p-9\right).

$$\left(6p^{2}-2p\right)+\left(27p-9\right)$$

Factor out 2p in the first and 9 in the second group.

$$2p\left(3p-1\right)+9\left(3p-1\right)$$

Factor out common term 3p-1 by using distributive property.

$$\left(3p-1\right)\left(2p+9\right)$$

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.

$$6p^{2}+25p-9=0$$

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.

$$p=\frac{-25±\sqrt{25^{2}-4\times 6\left(-9\right)}}{2\times 6}$$

Square 25.

$$p=\frac{-25±\sqrt{625-4\times 6\left(-9\right)}}{2\times 6}$$

Multiply -4 times 6.

$$p=\frac{-25±\sqrt{625-24\left(-9\right)}}{2\times 6}$$

Multiply -24 times -9.

$$p=\frac{-25±\sqrt{625+216}}{2\times 6}$$

Add 625 to 216.

$$p=\frac{-25±\sqrt{841}}{2\times 6}$$

Take the square root of 841.

$$p=\frac{-25±29}{2\times 6}$$

Multiply 2 times 6.

$$p=\frac{-25±29}{12}$$

Now solve the equation p=\frac{-25±29}{12} when ± is plus. Add -25 to 29.

$$p=\frac{4}{12}$$

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

$$p=\frac{1}{3}$$

Now solve the equation p=\frac{-25±29}{12} when ± is minus. Subtract 29 from -25.

$$p=-\frac{54}{12}$$

Reduce the fraction \frac{-54}{12} to lowest terms by extracting and canceling out 6.

$$p=-\frac{9}{2}$$

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 -\frac{9}{2} for x_{2}.

$$6p^{2}+25p-9=6\left(p-\frac{1}{3}\right)\left(p-\left(-\frac{9}{2}\right)\right)$$

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

$$6p^{2}+25p-9=6\left(p-\frac{1}{3}\right)\left(p+\frac{9}{2}\right)$$

Subtract \frac{1}{3} from p by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6p^{2}+25p-9=6\times \frac{3p-1}{3}\left(p+\frac{9}{2}\right)$$

Add \frac{9}{2} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6p^{2}+25p-9=6\times \frac{3p-1}{3}\times \frac{2p+9}{2}$$

Multiply \frac{3p-1}{3} times \frac{2p+9}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6p^{2}+25p-9=6\times \frac{\left(3p-1\right)\left(2p+9\right)}{3\times 2}$$

Multiply 3 times 2.

$$6p^{2}+25p-9=6\times \frac{\left(3p-1\right)\left(2p+9\right)}{6}$$

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

$$6p^{2}+25p-9=\left(3p-1\right)\left(2p+9\right)$$

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

$$x ^ 2 +\frac{25}{6}x -\frac{3}{2} = 0$$

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

$$r + s = -\frac{25}{6} rs = -\frac{3}{2}$$

Two numbers r and s sum up to -\frac{25}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{25}{6} = -\frac{25}{12}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{25}{12} - u s = -\frac{25}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{2}

$$(-\frac{25}{12} - u) (-\frac{25}{12} + u) = -\frac{3}{2}$$

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

$$\frac{625}{144} - u^2 = -\frac{3}{2}$$

Simplify the expression by subtracting \frac{625}{144} on both sides

$$-u^2 = -\frac{3}{2}-\frac{625}{144} = -\frac{841}{144}$$

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

$$u^2 = \frac{841}{144} u = \pm\sqrt{\frac{841}{144}} = \pm \frac{29}{12}$$

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

$$r =-\frac{25}{12} - \frac{29}{12} = -4.500 s = -\frac{25}{12} + \frac{29}{12} = 0.333$$

Evaluate

$$\left(3p-1\right)\left(2p+9\right)$$