Solve f(x)=-20x^2+23x-6?

$$f ( x ) = - 20 x ^ { 2 } + 23 x - 6 ?$$

Factor

$$\left(2-5x\right)\left(4x-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -20x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=23 ab=-20\left(-6\right)=120$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 23.

$$a=15 b=8$$

Rewrite -20x^{2}+23x-6 as \left(-20x^{2}+15x\right)+\left(8x-6\right).

$$\left(-20x^{2}+15x\right)+\left(8x-6\right)$$

Factor out -5x in the first and 2 in the second group.

$$-5x\left(4x-3\right)+2\left(4x-3\right)$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(-5x+2\right)$$

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.

$$-20x^{2}+23x-6=0$$

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.

$$x=\frac{-23±\sqrt{23^{2}-4\left(-20\right)\left(-6\right)}}{2\left(-20\right)}$$

Square 23.

$$x=\frac{-23±\sqrt{529-4\left(-20\right)\left(-6\right)}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-23±\sqrt{529+80\left(-6\right)}}{2\left(-20\right)}$$

Multiply 80 times -6.

$$x=\frac{-23±\sqrt{529-480}}{2\left(-20\right)}$$

Add 529 to -480.

$$x=\frac{-23±\sqrt{49}}{2\left(-20\right)}$$

Take the square root of 49.

$$x=\frac{-23±7}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{-23±7}{-40}$$

Now solve the equation x=\frac{-23±7}{-40} when ± is plus. Add -23 to 7.

$$x=-\frac{16}{-40}$$

Reduce the fraction \frac{-16}{-40} to lowest terms by extracting and canceling out 8.

$$x=\frac{2}{5}$$

Now solve the equation x=\frac{-23±7}{-40} when ± is minus. Subtract 7 from -23.

$$x=-\frac{30}{-40}$$

Reduce the fraction \frac{-30}{-40} to lowest terms by extracting and canceling out 10.

$$x=\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{5} for x_{1} and \frac{3}{4} for x_{2}.

$$-20x^{2}+23x-6=-20\left(x-\frac{2}{5}\right)\left(x-\frac{3}{4}\right)$$

Subtract \frac{2}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-20x^{2}+23x-6=-20\times \frac{-5x+2}{-5}\left(x-\frac{3}{4}\right)$$

Subtract \frac{3}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-20x^{2}+23x-6=-20\times \frac{-5x+2}{-5}\times \frac{-4x+3}{-4}$$

Multiply \frac{-5x+2}{-5} times \frac{-4x+3}{-4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-20x^{2}+23x-6=-20\times \frac{\left(-5x+2\right)\left(-4x+3\right)}{-5\left(-4\right)}$$

Multiply -5 times -4.

$$-20x^{2}+23x-6=-20\times \frac{\left(-5x+2\right)\left(-4x+3\right)}{20}$$

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

$$-20x^{2}+23x-6=-\left(-5x+2\right)\left(-4x+3\right)$$

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.

$$x ^ 2 -\frac{23}{20}x +\frac{3}{10} = 0$$

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

$$r + s = \frac{23}{20} rs = \frac{3}{10}$$

Two numbers r and s sum up to \frac{23}{20} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{20} = \frac{23}{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>

$$r = \frac{23}{40} - u s = \frac{23}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{10}

$$(\frac{23}{40} - u) (\frac{23}{40} + u) = \frac{3}{10}$$

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

$$\frac{529}{1600} - u^2 = \frac{3}{10}$$

Simplify the expression by subtracting \frac{529}{1600} on both sides

$$-u^2 = \frac{3}{10}-\frac{529}{1600} = -\frac{49}{1600}$$

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

$$u^2 = \frac{49}{1600} u = \pm\sqrt{\frac{49}{1600}} = \pm \frac{7}{40}$$

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

$$r =\frac{23}{40} - \frac{7}{40} = 0.400 s = \frac{23}{40} + \frac{7}{40} = 0.750$$

Evaluate

$$\left(2-5x\right)\left(4x-3\right)$$