Solve 2v^2-23v+63

$$2 v ^ { 2 } - 23 v + 63$$

Factor

$$\left(v-7\right)\left(2v-9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2v^{2}+av+bv+63. To find a and b, set up a system to be solved.

$$a+b=-23 ab=2\times 63=126$$

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 126.

$$-1,-126 -2,-63 -3,-42 -6,-21 -7,-18 -9,-14$$

Calculate the sum for each pair.

$$-1-126=-127 -2-63=-65 -3-42=-45 -6-21=-27 -7-18=-25 -9-14=-23$$

The solution is the pair that gives sum -23.

$$a=-14 b=-9$$

Rewrite 2v^{2}-23v+63 as \left(2v^{2}-14v\right)+\left(-9v+63\right).

$$\left(2v^{2}-14v\right)+\left(-9v+63\right)$$

Factor out 2v in the first and -9 in the second group.

$$2v\left(v-7\right)-9\left(v-7\right)$$

Factor out common term v-7 by using distributive property.

$$\left(v-7\right)\left(2v-9\right)$$

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.

$$2v^{2}-23v+63=0$$

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.

$$v=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 2\times 63}}{2\times 2}$$

Square -23.

$$v=\frac{-\left(-23\right)±\sqrt{529-4\times 2\times 63}}{2\times 2}$$

Multiply -4 times 2.

$$v=\frac{-\left(-23\right)±\sqrt{529-8\times 63}}{2\times 2}$$

Multiply -8 times 63.

$$v=\frac{-\left(-23\right)±\sqrt{529-504}}{2\times 2}$$

Add 529 to -504.

$$v=\frac{-\left(-23\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$v=\frac{-\left(-23\right)±5}{2\times 2}$$

The opposite of -23 is 23.

$$v=\frac{23±5}{2\times 2}$$

Multiply 2 times 2.

$$v=\frac{23±5}{4}$$

Now solve the equation v=\frac{23±5}{4} when ± is plus. Add 23 to 5.

$$v=\frac{28}{4}$$

Divide 28 by 4.

$$v=7$$

Now solve the equation v=\frac{23±5}{4} when ± is minus. Subtract 5 from 23.

$$v=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$v=\frac{9}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 7 for x_{1} and \frac{9}{2} for x_{2}.

$$2v^{2}-23v+63=2\left(v-7\right)\left(v-\frac{9}{2}\right)$$

Subtract \frac{9}{2} from v by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2v^{2}-23v+63=2\left(v-7\right)\times \frac{2v-9}{2}$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2v^{2}-23v+63=\left(v-7\right)\left(2v-9\right)$$

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 2

$$x ^ 2 -\frac{23}{2}x +\frac{63}{2} = 0$$

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

$$r + s = \frac{23}{2} rs = \frac{63}{2}$$

Two numbers r and s sum up to \frac{23}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{2} = \frac{23}{4}. 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>

$$r = \frac{23}{4} - u s = \frac{23}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{63}{2}

$$(\frac{23}{4} - u) (\frac{23}{4} + u) = \frac{63}{2}$$

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

$$\frac{529}{16} - u^2 = \frac{63}{2}$$

Simplify the expression by subtracting \frac{529}{16} on both sides

$$-u^2 = \frac{63}{2}-\frac{529}{16} = -\frac{25}{16}$$

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

$$u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{4}$$

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

$$r =\frac{23}{4} - \frac{5}{4} = 4.500 s = \frac{23}{4} + \frac{5}{4} = 7$$

Evaluate

$$\left(v-7\right)\left(2v-9\right)$$