Solve -3v^2-v+14

$$- 3 v ^ { 2 } - v + 14$$

Factor

$$\left(2-v\right)\left(3v+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -3v^{2}+av+bv+14. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-3\times 14=-42$$

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 -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -1.

$$a=6 b=-7$$

Rewrite -3v^{2}-v+14 as \left(-3v^{2}+6v\right)+\left(-7v+14\right).

$$\left(-3v^{2}+6v\right)+\left(-7v+14\right)$$

Factor out 3v in the first and 7 in the second group.

$$3v\left(-v+2\right)+7\left(-v+2\right)$$

Factor out common term -v+2 by using distributive property.

$$\left(-v+2\right)\left(3v+7\right)$$

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.

$$-3v^{2}-v+14=0$$

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.

$$v=\frac{-\left(-1\right)±\sqrt{1-4\left(-3\right)\times 14}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$v=\frac{-\left(-1\right)±\sqrt{1+12\times 14}}{2\left(-3\right)}$$

Multiply 12 times 14.

$$v=\frac{-\left(-1\right)±\sqrt{1+168}}{2\left(-3\right)}$$

Add 1 to 168.

$$v=\frac{-\left(-1\right)±\sqrt{169}}{2\left(-3\right)}$$

Take the square root of 169.

$$v=\frac{-\left(-1\right)±13}{2\left(-3\right)}$$

The opposite of -1 is 1.

$$v=\frac{1±13}{2\left(-3\right)}$$

Multiply 2 times -3.

$$v=\frac{1±13}{-6}$$

Now solve the equation v=\frac{1±13}{-6} when ± is plus. Add 1 to 13.

$$v=\frac{14}{-6}$$

Reduce the fraction \frac{14}{-6} to lowest terms by extracting and canceling out 2.

$$v=-\frac{7}{3}$$

Now solve the equation v=\frac{1±13}{-6} when ± is minus. Subtract 13 from 1.

$$v=-\frac{12}{-6}$$

Divide -12 by -6.

$$v=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{7}{3} for x_{1} and 2 for x_{2}.

$$-3v^{2}-v+14=-3\left(v-\left(-\frac{7}{3}\right)\right)\left(v-2\right)$$

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

$$-3v^{2}-v+14=-3\left(v+\frac{7}{3}\right)\left(v-2\right)$$

Add \frac{7}{3} to v by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-3v^{2}-v+14=-3\times \frac{-3v-7}{-3}\left(v-2\right)$$

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

$$-3v^{2}-v+14=\left(-3v-7\right)\left(v-2\right)$$

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.

$$x ^ 2 +\frac{1}{3}x -\frac{14}{3} = 0$$

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

$$r + s = -\frac{1}{3} rs = -\frac{14}{3}$$

Two numbers r and s sum up to -\frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{3} = -\frac{1}{6}. 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>

$$r = -\frac{1}{6} - u s = -\frac{1}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{14}{3}

$$(-\frac{1}{6} - u) (-\frac{1}{6} + u) = -\frac{14}{3}$$

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

$$\frac{1}{36} - u^2 = -\frac{14}{3}$$

Simplify the expression by subtracting \frac{1}{36} on both sides

$$-u^2 = -\frac{14}{3}-\frac{1}{36} = -\frac{169}{36}$$

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

$$u^2 = \frac{169}{36} u = \pm\sqrt{\frac{169}{36}} = \pm \frac{13}{6}$$

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

$$r =-\frac{1}{6} - \frac{13}{6} = -2.333 s = -\frac{1}{6} + \frac{13}{6} = 2.000$$

Evaluate

$$\left(2-v\right)\left(3v+7\right)$$