Solve -3p^2-8p-5

$$- 3 p ^ { 2 } - 8 p - 5$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -3p^{2}+ap+bp-5. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-3\left(-5\right)=15$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-3 b=-5$$

Rewrite -3p^{2}-8p-5 as \left(-3p^{2}-3p\right)+\left(-5p-5\right).

$$\left(-3p^{2}-3p\right)+\left(-5p-5\right)$$

Factor out 3p in the first and 5 in the second group.

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

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

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

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.

$$-3p^{2}-8p-5=0$$

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.

$$p=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}$$

Square -8.

$$p=\frac{-\left(-8\right)±\sqrt{64-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$p=\frac{-\left(-8\right)±\sqrt{64+12\left(-5\right)}}{2\left(-3\right)}$$

Multiply 12 times -5.

$$p=\frac{-\left(-8\right)±\sqrt{64-60}}{2\left(-3\right)}$$

Add 64 to -60.

$$p=\frac{-\left(-8\right)±\sqrt{4}}{2\left(-3\right)}$$

Take the square root of 4.

$$p=\frac{-\left(-8\right)±2}{2\left(-3\right)}$$

The opposite of -8 is 8.

$$p=\frac{8±2}{2\left(-3\right)}$$

Multiply 2 times -3.

$$p=\frac{8±2}{-6}$$

Now solve the equation p=\frac{8±2}{-6} when ± is plus. Add 8 to 2.

$$p=\frac{10}{-6}$$

Reduce the fraction \frac{10}{-6} to lowest terms by extracting and canceling out 2.

$$p=-\frac{5}{3}$$

Now solve the equation p=\frac{8±2}{-6} when ± is minus. Subtract 2 from 8.

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

Divide 6 by -6.

$$p=-1$$

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

$$-3p^{2}-8p-5=-3\left(p-\left(-\frac{5}{3}\right)\right)\left(p-\left(-1\right)\right)$$

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

$$-3p^{2}-8p-5=-3\left(p+\frac{5}{3}\right)\left(p+1\right)$$

Add \frac{5}{3} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-3p^{2}-8p-5=-3\times \frac{-3p-5}{-3}\left(p+1\right)$$

Cancel out 3, the greatest common factor in -3 and 3.

$$-3p^{2}-8p-5=\left(-3p-5\right)\left(p+1\right)$$

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.

$$x ^ 2 +\frac{8}{3}x +\frac{5}{3} = 0$$

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

$$r + s = -\frac{8}{3} rs = \frac{5}{3}$$

Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{3}. 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>

$$r = -\frac{4}{3} - u s = -\frac{4}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{3}

$$(-\frac{4}{3} - u) (-\frac{4}{3} + u) = \frac{5}{3}$$

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

$$\frac{16}{9} - u^2 = \frac{5}{3}$$

Simplify the expression by subtracting \frac{16}{9} on both sides

$$-u^2 = \frac{5}{3}-\frac{16}{9} = -\frac{1}{9}$$

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

$$u^2 = \frac{1}{9} u = \pm\sqrt{\frac{1}{9}} = \pm \frac{1}{3}$$

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

$$r =-\frac{4}{3} - \frac{1}{3} = -1.667 s = -\frac{4}{3} + \frac{1}{3} = -1.000$$

Evaluate

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