Solve v^2+3v-28

$$v ^ { 2 } + 3 v - 28$$

Factor

$$\left(v-4\right)\left(v+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as v^{2}+av+bv-28. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-28\right)=-28$$

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 -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=-4 b=7$$

Rewrite v^{2}+3v-28 as \left(v^{2}-4v\right)+\left(7v-28\right).

$$\left(v^{2}-4v\right)+\left(7v-28\right)$$

Factor out v in the first and 7 in the second group.

$$v\left(v-4\right)+7\left(v-4\right)$$

Factor out common term v-4 by using distributive property.

$$\left(v-4\right)\left(v+7\right)$$

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.

$$v^{2}+3v-28=0$$

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.

$$v=\frac{-3±\sqrt{3^{2}-4\left(-28\right)}}{2}$$

Square 3.

$$v=\frac{-3±\sqrt{9-4\left(-28\right)}}{2}$$

Multiply -4 times -28.

$$v=\frac{-3±\sqrt{9+112}}{2}$$

Add 9 to 112.

$$v=\frac{-3±\sqrt{121}}{2}$$

Take the square root of 121.

$$v=\frac{-3±11}{2}$$

Now solve the equation v=\frac{-3±11}{2} when ± is plus. Add -3 to 11.

$$v=\frac{8}{2}$$

Divide 8 by 2.

$$v=4$$

Now solve the equation v=\frac{-3±11}{2} when ± is minus. Subtract 11 from -3.

$$v=-\frac{14}{2}$$

Divide -14 by 2.

$$v=-7$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and -7 for x_{2}.

$$v^{2}+3v-28=\left(v-4\right)\left(v-\left(-7\right)\right)$$

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

$$v^{2}+3v-28=\left(v-4\right)\left(v+7\right)$$

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.

$$x ^ 2 +3x -28 = 0$$

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

$$r + s = -3 rs = -28$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -28

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -28$$

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

$$\frac{9}{4} - u^2 = -28$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -28-\frac{9}{4} = -\frac{121}{4}$$

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

$$u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{2}$$

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

$$r =-\frac{3}{2} - \frac{11}{2} = -7 s = -\frac{3}{2} + \frac{11}{2} = 4$$

Evaluate

$$\left(v-4\right)\left(v+7\right)$$