Solve 15p^2+7p-2

$$15 p ^ { 2 } + 7 p - 2$$

Factor

$$\left(5p-1\right)\left(3p+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15p^{2}+ap+bp-2. To find a and b, set up a system to be solved.

$$a+b=7 ab=15\left(-2\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=-3 b=10$$

Rewrite 15p^{2}+7p-2 as \left(15p^{2}-3p\right)+\left(10p-2\right).

$$\left(15p^{2}-3p\right)+\left(10p-2\right)$$

Factor out 3p in the first and 2 in the second group.

$$3p\left(5p-1\right)+2\left(5p-1\right)$$

Factor out common term 5p-1 by using distributive property.

$$\left(5p-1\right)\left(3p+2\right)$$

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.

$$15p^{2}+7p-2=0$$

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.

$$p=\frac{-7±\sqrt{7^{2}-4\times 15\left(-2\right)}}{2\times 15}$$

Square 7.

$$p=\frac{-7±\sqrt{49-4\times 15\left(-2\right)}}{2\times 15}$$

Multiply -4 times 15.

$$p=\frac{-7±\sqrt{49-60\left(-2\right)}}{2\times 15}$$

Multiply -60 times -2.

$$p=\frac{-7±\sqrt{49+120}}{2\times 15}$$

Add 49 to 120.

$$p=\frac{-7±\sqrt{169}}{2\times 15}$$

Take the square root of 169.

$$p=\frac{-7±13}{2\times 15}$$

Multiply 2 times 15.

$$p=\frac{-7±13}{30}$$

Now solve the equation p=\frac{-7±13}{30} when ± is plus. Add -7 to 13.

$$p=\frac{6}{30}$$

Reduce the fraction \frac{6}{30} to lowest terms by extracting and canceling out 6.

$$p=\frac{1}{5}$$

Now solve the equation p=\frac{-7±13}{30} when ± is minus. Subtract 13 from -7.

$$p=-\frac{20}{30}$$

Reduce the fraction \frac{-20}{30} to lowest terms by extracting and canceling out 10.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{5} for x_{1} and -\frac{2}{3} for x_{2}.

$$15p^{2}+7p-2=15\left(p-\frac{1}{5}\right)\left(p-\left(-\frac{2}{3}\right)\right)$$

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

$$15p^{2}+7p-2=15\left(p-\frac{1}{5}\right)\left(p+\frac{2}{3}\right)$$

Subtract \frac{1}{5} from p by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15p^{2}+7p-2=15\times \frac{5p-1}{5}\left(p+\frac{2}{3}\right)$$

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

$$15p^{2}+7p-2=15\times \frac{5p-1}{5}\times \frac{3p+2}{3}$$

Multiply \frac{5p-1}{5} times \frac{3p+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15p^{2}+7p-2=15\times \frac{\left(5p-1\right)\left(3p+2\right)}{5\times 3}$$

Multiply 5 times 3.

$$15p^{2}+7p-2=15\times \frac{\left(5p-1\right)\left(3p+2\right)}{15}$$

Cancel out 15, the greatest common factor in 15 and 15.

$$15p^{2}+7p-2=\left(5p-1\right)\left(3p+2\right)$$

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 15

$$x ^ 2 +\frac{7}{15}x -\frac{2}{15} = 0$$

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

$$r + s = -\frac{7}{15} rs = -\frac{2}{15}$$

Two numbers r and s sum up to -\frac{7}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{15} = -\frac{7}{30}. 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>

$$r = -\frac{7}{30} - u s = -\frac{7}{30} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{15}

$$(-\frac{7}{30} - u) (-\frac{7}{30} + u) = -\frac{2}{15}$$

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

$$\frac{49}{900} - u^2 = -\frac{2}{15}$$

Simplify the expression by subtracting \frac{49}{900} on both sides

$$-u^2 = -\frac{2}{15}-\frac{49}{900} = -\frac{169}{900}$$

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

$$u^2 = \frac{169}{900} u = \pm\sqrt{\frac{169}{900}} = \pm \frac{13}{30}$$

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

$$r =-\frac{7}{30} - \frac{13}{30} = -0.667 s = -\frac{7}{30} + \frac{13}{30} = 0.200$$

Evaluate

$$\left(5p-1\right)\left(3p+2\right)$$