Solve 18p^2-25p-3

$$18 p ^ { 2 } - 25 p - 3$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 18p^{2}+ap+bp-3. To find a and b, set up a system to be solved.

$$a+b=-25 ab=18\left(-3\right)=-54$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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.

$$a=-27 b=2$$

Rewrite 18p^{2}-25p-3 as \left(18p^{2}-27p\right)+\left(2p-3\right).

$$\left(18p^{2}-27p\right)+\left(2p-3\right)$$

Factor out 9p in 18p^{2}-27p.

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

Factor out common term 2p-3 by using distributive property.

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

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.

$$18p^{2}-25p-3=0$$

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.

$$p=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 18\left(-3\right)}}{2\times 18}$$

Square -25.

$$p=\frac{-\left(-25\right)±\sqrt{625-4\times 18\left(-3\right)}}{2\times 18}$$

Multiply -4 times 18.

$$p=\frac{-\left(-25\right)±\sqrt{625-72\left(-3\right)}}{2\times 18}$$

Multiply -72 times -3.

$$p=\frac{-\left(-25\right)±\sqrt{625+216}}{2\times 18}$$

Add 625 to 216.

$$p=\frac{-\left(-25\right)±\sqrt{841}}{2\times 18}$$

Take the square root of 841.

$$p=\frac{-\left(-25\right)±29}{2\times 18}$$

The opposite of -25 is 25.

$$p=\frac{25±29}{2\times 18}$$

Multiply 2 times 18.

$$p=\frac{25±29}{36}$$

Now solve the equation p=\frac{25±29}{36} when ± is plus. Add 25 to 29.

$$p=\frac{54}{36}$$

Reduce the fraction \frac{54}{36} to lowest terms by extracting and canceling out 18.

$$p=\frac{3}{2}$$

Now solve the equation p=\frac{25±29}{36} when ± is minus. Subtract 29 from 25.

$$p=-\frac{4}{36}$$

Reduce the fraction \frac{-4}{36} to lowest terms by extracting and canceling out 4.

$$p=-\frac{1}{9}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and -\frac{1}{9} for x_{2}.

$$18p^{2}-25p-3=18\left(p-\frac{3}{2}\right)\left(p-\left(-\frac{1}{9}\right)\right)$$

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

$$18p^{2}-25p-3=18\left(p-\frac{3}{2}\right)\left(p+\frac{1}{9}\right)$$

Subtract \frac{3}{2} from p by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18p^{2}-25p-3=18\times \frac{2p-3}{2}\left(p+\frac{1}{9}\right)$$

Add \frac{1}{9} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18p^{2}-25p-3=18\times \frac{2p-3}{2}\times \frac{9p+1}{9}$$

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

$$18p^{2}-25p-3=18\times \frac{\left(2p-3\right)\left(9p+1\right)}{2\times 9}$$

Multiply 2 times 9.

$$18p^{2}-25p-3=18\times \frac{\left(2p-3\right)\left(9p+1\right)}{18}$$

Cancel out 18, the greatest common factor in 18 and 18.

$$18p^{2}-25p-3=\left(2p-3\right)\left(9p+1\right)$$

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 18

$$x ^ 2 -\frac{25}{18}x -\frac{1}{6} = 0$$

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

$$r + s = \frac{25}{18} rs = -\frac{1}{6}$$

Two numbers r and s sum up to \frac{25}{18} exactly when the average of the two numbers is \frac{1}{2}*\frac{25}{18} = \frac{25}{36}. 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>

$$r = \frac{25}{36} - u s = \frac{25}{36} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{6}

$$(\frac{25}{36} - u) (\frac{25}{36} + u) = -\frac{1}{6}$$

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

$$\frac{625}{1296} - u^2 = -\frac{1}{6}$$

Simplify the expression by subtracting \frac{625}{1296} on both sides

$$-u^2 = -\frac{1}{6}-\frac{625}{1296} = -\frac{841}{1296}$$

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

$$u^2 = \frac{841}{1296} u = \pm\sqrt{\frac{841}{1296}} = \pm \frac{29}{36}$$

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

$$r =\frac{25}{36} - \frac{29}{36} = -0.111 s = \frac{25}{36} + \frac{29}{36} = 1.500$$

Evaluate

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