Solve -3x^2+17x-20=

$$- 3 x ^ { 2 } + 17 x - 20 =$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=17 ab=-3\left(-20\right)=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 17.

$$a=12 b=5$$

Rewrite -3x^{2}+17x-20 as \left(-3x^{2}+12x\right)+\left(5x-20\right).

$$\left(-3x^{2}+12x\right)+\left(5x-20\right)$$

Factor out 3x in the first and -5 in the second group.

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

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

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

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.

$$-3x^{2}+17x-20=0$$

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.

$$x=\frac{-17±\sqrt{17^{2}-4\left(-3\right)\left(-20\right)}}{2\left(-3\right)}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\left(-3\right)\left(-20\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-17±\sqrt{289+12\left(-20\right)}}{2\left(-3\right)}$$

Multiply 12 times -20.

$$x=\frac{-17±\sqrt{289-240}}{2\left(-3\right)}$$

Add 289 to -240.

$$x=\frac{-17±\sqrt{49}}{2\left(-3\right)}$$

Take the square root of 49.

$$x=\frac{-17±7}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-17±7}{-6}$$

Now solve the equation x=\frac{-17±7}{-6} when ± is plus. Add -17 to 7.

$$x=-\frac{10}{-6}$$

Reduce the fraction \frac{-10}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{-17±7}{-6} when ± is minus. Subtract 7 from -17.

$$x=-\frac{24}{-6}$$

Divide -24 by -6.

$$x=4$$

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

$$-3x^{2}+17x-20=-3\left(x-\frac{5}{3}\right)\left(x-4\right)$$

Subtract \frac{5}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-3x^{2}+17x-20=-3\times \frac{-3x+5}{-3}\left(x-4\right)$$

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

$$-3x^{2}+17x-20=\left(-3x+5\right)\left(x-4\right)$$

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.

$$x ^ 2 -\frac{17}{3}x +\frac{20}{3} = 0$$

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

$$r + s = \frac{17}{3} rs = \frac{20}{3}$$

Two numbers r and s sum up to \frac{17}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{3} = \frac{17}{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>

$$r = \frac{17}{6} - u s = \frac{17}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{20}{3}

$$(\frac{17}{6} - u) (\frac{17}{6} + u) = \frac{20}{3}$$

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

$$\frac{289}{36} - u^2 = \frac{20}{3}$$

Simplify the expression by subtracting \frac{289}{36} on both sides

$$-u^2 = \frac{20}{3}-\frac{289}{36} = -\frac{49}{36}$$

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

$$u^2 = \frac{49}{36} u = \pm\sqrt{\frac{49}{36}} = \pm \frac{7}{6}$$

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

$$r =\frac{17}{6} - \frac{7}{6} = 1.667 s = \frac{17}{6} + \frac{7}{6} = 4$$

Evaluate

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