Solve 20x^2-17x-63

$$20 x ^ { 2 } - 17 x - 63$$

Factor

$$\left(4x-9\right)\left(5x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx-63. To find a and b, set up a system to be solved.

$$a+b=-17 ab=20\left(-63\right)=-1260$$

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 -1260.

$$1,-1260 2,-630 3,-420 4,-315 5,-252 6,-210 7,-180 9,-140 10,-126 12,-105 14,-90 15,-84 18,-70 20,-63 21,-60 28,-45 30,-42 35,-36$$

Calculate the sum for each pair.

$$1-1260=-1259 2-630=-628 3-420=-417 4-315=-311 5-252=-247 6-210=-204 7-180=-173 9-140=-131 10-126=-116 12-105=-93 14-90=-76 15-84=-69 18-70=-52 20-63=-43 21-60=-39 28-45=-17 30-42=-12 35-36=-1$$

The solution is the pair that gives sum -17.

$$a=-45 b=28$$

Rewrite 20x^{2}-17x-63 as \left(20x^{2}-45x\right)+\left(28x-63\right).

$$\left(20x^{2}-45x\right)+\left(28x-63\right)$$

Factor out 5x in the first and 7 in the second group.

$$5x\left(4x-9\right)+7\left(4x-9\right)$$

Factor out common term 4x-9 by using distributive property.

$$\left(4x-9\right)\left(5x+7\right)$$

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.

$$20x^{2}-17x-63=0$$

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.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 20\left(-63\right)}}{2\times 20}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 20\left(-63\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-17\right)±\sqrt{289-80\left(-63\right)}}{2\times 20}$$

Multiply -80 times -63.

$$x=\frac{-\left(-17\right)±\sqrt{289+5040}}{2\times 20}$$

Add 289 to 5040.

$$x=\frac{-\left(-17\right)±\sqrt{5329}}{2\times 20}$$

Take the square root of 5329.

$$x=\frac{-\left(-17\right)±73}{2\times 20}$$

The opposite of -17 is 17.

$$x=\frac{17±73}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{17±73}{40}$$

Now solve the equation x=\frac{17±73}{40} when ± is plus. Add 17 to 73.

$$x=\frac{90}{40}$$

Reduce the fraction \frac{90}{40} to lowest terms by extracting and canceling out 10.

$$x=\frac{9}{4}$$

Now solve the equation x=\frac{17±73}{40} when ± is minus. Subtract 73 from 17.

$$x=-\frac{56}{40}$$

Reduce the fraction \frac{-56}{40} to lowest terms by extracting and canceling out 8.

$$x=-\frac{7}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{4} for x_{1} and -\frac{7}{5} for x_{2}.

$$20x^{2}-17x-63=20\left(x-\frac{9}{4}\right)\left(x-\left(-\frac{7}{5}\right)\right)$$

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

$$20x^{2}-17x-63=20\left(x-\frac{9}{4}\right)\left(x+\frac{7}{5}\right)$$

Subtract \frac{9}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-17x-63=20\times \frac{4x-9}{4}\left(x+\frac{7}{5}\right)$$

Add \frac{7}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-17x-63=20\times \frac{4x-9}{4}\times \frac{5x+7}{5}$$

Multiply \frac{4x-9}{4} times \frac{5x+7}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-17x-63=20\times \frac{\left(4x-9\right)\left(5x+7\right)}{4\times 5}$$

Multiply 4 times 5.

$$20x^{2}-17x-63=20\times \frac{\left(4x-9\right)\left(5x+7\right)}{20}$$

Cancel out 20, the greatest common factor in 20 and 20.

$$20x^{2}-17x-63=\left(4x-9\right)\left(5x+7\right)$$

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.This is achieved by dividing both sides of the equation by 20

$$x ^ 2 -\frac{17}{20}x -\frac{63}{20} = 0$$

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

$$r + s = \frac{17}{20} rs = -\frac{63}{20}$$

Two numbers r and s sum up to \frac{17}{20} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{20} = \frac{17}{40}. 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>

$$r = \frac{17}{40} - u s = \frac{17}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{63}{20}

$$(\frac{17}{40} - u) (\frac{17}{40} + u) = -\frac{63}{20}$$

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

$$\frac{289}{1600} - u^2 = -\frac{63}{20}$$

Simplify the expression by subtracting \frac{289}{1600} on both sides

$$-u^2 = -\frac{63}{20}-\frac{289}{1600} = -\frac{5329}{1600}$$

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

$$u^2 = \frac{5329}{1600} u = \pm\sqrt{\frac{5329}{1600}} = \pm \frac{73}{40}$$

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

$$r =\frac{17}{40} - \frac{73}{40} = -1.400 s = \frac{17}{40} + \frac{73}{40} = 2.250$$

Evaluate

$$\left(4x-9\right)\left(5x+7\right)$$