Solve -5v^2-28v-15

$$- 5 v ^ { 2 } - 28 v - 15$$

Factor

$$-\left(v+5\right)\left(5v+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -5v^{2}+av+bv-15. To find a and b, set up a system to be solved.

$$a+b=-28 ab=-5\left(-15\right)=75$$

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 75.

$$-1,-75 -3,-25 -5,-15$$

Calculate the sum for each pair.

$$-1-75=-76 -3-25=-28 -5-15=-20$$

The solution is the pair that gives sum -28.

$$a=-3 b=-25$$

Rewrite -5v^{2}-28v-15 as \left(-5v^{2}-3v\right)+\left(-25v-15\right).

$$\left(-5v^{2}-3v\right)+\left(-25v-15\right)$$

Factor out -v in the first and -5 in the second group.

$$-v\left(5v+3\right)-5\left(5v+3\right)$$

Factor out common term 5v+3 by using distributive property.

$$\left(5v+3\right)\left(-v-5\right)$$

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.

$$-5v^{2}-28v-15=0$$

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.

$$v=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\left(-5\right)\left(-15\right)}}{2\left(-5\right)}$$

Square -28.

$$v=\frac{-\left(-28\right)±\sqrt{784-4\left(-5\right)\left(-15\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$v=\frac{-\left(-28\right)±\sqrt{784+20\left(-15\right)}}{2\left(-5\right)}$$

Multiply 20 times -15.

$$v=\frac{-\left(-28\right)±\sqrt{784-300}}{2\left(-5\right)}$$

Add 784 to -300.

$$v=\frac{-\left(-28\right)±\sqrt{484}}{2\left(-5\right)}$$

Take the square root of 484.

$$v=\frac{-\left(-28\right)±22}{2\left(-5\right)}$$

The opposite of -28 is 28.

$$v=\frac{28±22}{2\left(-5\right)}$$

Multiply 2 times -5.

$$v=\frac{28±22}{-10}$$

Now solve the equation v=\frac{28±22}{-10} when ± is plus. Add 28 to 22.

$$v=\frac{50}{-10}$$

Divide 50 by -10.

$$v=-5$$

Now solve the equation v=\frac{28±22}{-10} when ± is minus. Subtract 22 from 28.

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

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

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

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

$$-5v^{2}-28v-15=-5\left(v-\left(-5\right)\right)\left(v-\left(-\frac{3}{5}\right)\right)$$

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

$$-5v^{2}-28v-15=-5\left(v+5\right)\left(v+\frac{3}{5}\right)$$

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

$$-5v^{2}-28v-15=-5\left(v+5\right)\times \frac{-5v-3}{-5}$$

Cancel out 5, the greatest common factor in -5 and 5.

$$-5v^{2}-28v-15=\left(v+5\right)\left(-5v-3\right)$$

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.

$$x ^ 2 +\frac{28}{5}x +3 = 0$$

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

$$r + s = -\frac{28}{5} rs = 3$$

Two numbers r and s sum up to -\frac{28}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{28}{5} = -\frac{14}{5}. 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>

$$r = -\frac{14}{5} - u s = -\frac{14}{5} + u$$

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

$$(-\frac{14}{5} - u) (-\frac{14}{5} + u) = 3$$

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

$$\frac{196}{25} - u^2 = 3$$

Simplify the expression by subtracting \frac{196}{25} on both sides

$$-u^2 = 3-\frac{196}{25} = -\frac{121}{25}$$

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

$$u^2 = \frac{121}{25} u = \pm\sqrt{\frac{121}{25}} = \pm \frac{11}{5}$$

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

$$r =-\frac{14}{5} - \frac{11}{5} = -5 s = -\frac{14}{5} + \frac{11}{5} = -0.600$$

Evaluate

$$-\left(v+5\right)\left(5v+3\right)$$