Solve 10p^2-55p-30

$$10 p ^ { 2 } - 55 p - 30$$

Factor

$$5\left(p-6\right)\left(2p+1\right)$$

Solution Steps

Factor out 5.

$$5\left(2p^{2}-11p-6\right)$$

Consider 2p^{2}-11p-6. Factor the expression by grouping. First, the expression needs to be rewritten as 2p^{2}+ap+bp-6. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\left(-6\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -11.

$$a=-12 b=1$$

Rewrite 2p^{2}-11p-6 as \left(2p^{2}-12p\right)+\left(p-6\right).

$$\left(2p^{2}-12p\right)+\left(p-6\right)$$

Factor out 2p in 2p^{2}-12p.

$$2p\left(p-6\right)+p-6$$

Factor out common term p-6 by using distributive property.

$$\left(p-6\right)\left(2p+1\right)$$

Rewrite the complete factored expression.

$$5\left(p-6\right)\left(2p+1\right)$$

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.

$$10p^{2}-55p-30=0$$

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.

$$p=\frac{-\left(-55\right)±\sqrt{\left(-55\right)^{2}-4\times 10\left(-30\right)}}{2\times 10}$$

Square -55.

$$p=\frac{-\left(-55\right)±\sqrt{3025-4\times 10\left(-30\right)}}{2\times 10}$$

Multiply -4 times 10.

$$p=\frac{-\left(-55\right)±\sqrt{3025-40\left(-30\right)}}{2\times 10}$$

Multiply -40 times -30.

$$p=\frac{-\left(-55\right)±\sqrt{3025+1200}}{2\times 10}$$

Add 3025 to 1200.

$$p=\frac{-\left(-55\right)±\sqrt{4225}}{2\times 10}$$

Take the square root of 4225.

$$p=\frac{-\left(-55\right)±65}{2\times 10}$$

The opposite of -55 is 55.

$$p=\frac{55±65}{2\times 10}$$

Multiply 2 times 10.

$$p=\frac{55±65}{20}$$

Now solve the equation p=\frac{55±65}{20} when ± is plus. Add 55 to 65.

$$p=\frac{120}{20}$$

Divide 120 by 20.

$$p=6$$

Now solve the equation p=\frac{55±65}{20} when ± is minus. Subtract 65 from 55.

$$p=-\frac{10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

$$p=-\frac{1}{2}$$

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

$$10p^{2}-55p-30=10\left(p-6\right)\left(p-\left(-\frac{1}{2}\right)\right)$$

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

$$10p^{2}-55p-30=10\left(p-6\right)\left(p+\frac{1}{2}\right)$$

Add \frac{1}{2} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10p^{2}-55p-30=10\left(p-6\right)\times \frac{2p+1}{2}$$

Cancel out 2, the greatest common factor in 10 and 2.

$$10p^{2}-55p-30=5\left(p-6\right)\left(2p+1\right)$$

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 10

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

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

$$r + s = \frac{11}{2} rs = -3$$

Two numbers r and s sum up to \frac{11}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{2} = \frac{11}{4}. 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>

$$r = \frac{11}{4} - u s = \frac{11}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -3

$$(\frac{11}{4} - u) (\frac{11}{4} + u) = -3$$

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

$$\frac{121}{16} - u^2 = -3$$

Simplify the expression by subtracting \frac{121}{16} on both sides

$$-u^2 = -3-\frac{121}{16} = -\frac{169}{16}$$

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

$$u^2 = \frac{169}{16} u = \pm\sqrt{\frac{169}{16}} = \pm \frac{13}{4}$$

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

$$r =\frac{11}{4} - \frac{13}{4} = -0.500 s = \frac{11}{4} + \frac{13}{4} = 6$$

Evaluate

$$5\left(p-6\right)\left(2p+1\right)$$