Solve 20s^2+17s-63

$$20 s ^ { 2 } + 17 s - 63$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20s^{2}+as+bs-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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-28 b=45$$

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

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

Factor out 4s in the first and 9 in the second group.

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

Factor out common term 5s-7 by using distributive property.

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

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.

$$20s^{2}+17s-63=0$$

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.

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

Square 17.

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

Multiply -4 times 20.

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

Multiply -80 times -63.

$$s=\frac{-17±\sqrt{289+5040}}{2\times 20}$$

Add 289 to 5040.

$$s=\frac{-17±\sqrt{5329}}{2\times 20}$$

Take the square root of 5329.

$$s=\frac{-17±73}{2\times 20}$$

Multiply 2 times 20.

$$s=\frac{-17±73}{40}$$

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

$$s=\frac{56}{40}$$

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

$$s=\frac{7}{5}$$

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

$$s=-\frac{90}{40}$$

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

$$s=-\frac{9}{4}$$

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

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

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

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

Subtract \frac{7}{5} from s by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Add \frac{9}{4} to s by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Multiply 5 times 4.

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

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

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

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 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.

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

Evaluate

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