Solve g(x)=-2x^2+29x-90

$$g ( x ) = - 2 x ^ { 2 } + 29 x - 90$$

Factor

$$\left(10-x\right)\left(2x-9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx-90. To find a and b, set up a system to be solved.

$$a+b=29 ab=-2\left(-90\right)=180$$

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 180.

$$1,180 2,90 3,60 4,45 5,36 6,30 9,20 10,18 12,15$$

Calculate the sum for each pair.

$$1+180=181 2+90=92 3+60=63 4+45=49 5+36=41 6+30=36 9+20=29 10+18=28 12+15=27$$

The solution is the pair that gives sum 29.

$$a=20 b=9$$

Rewrite -2x^{2}+29x-90 as \left(-2x^{2}+20x\right)+\left(9x-90\right).

$$\left(-2x^{2}+20x\right)+\left(9x-90\right)$$

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

$$2x\left(-x+10\right)-9\left(-x+10\right)$$

Factor out common term -x+10 by using distributive property.

$$\left(-x+10\right)\left(2x-9\right)$$

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.

$$-2x^{2}+29x-90=0$$

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.

$$x=\frac{-29±\sqrt{29^{2}-4\left(-2\right)\left(-90\right)}}{2\left(-2\right)}$$

Square 29.

$$x=\frac{-29±\sqrt{841-4\left(-2\right)\left(-90\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-29±\sqrt{841+8\left(-90\right)}}{2\left(-2\right)}$$

Multiply 8 times -90.

$$x=\frac{-29±\sqrt{841-720}}{2\left(-2\right)}$$

Add 841 to -720.

$$x=\frac{-29±\sqrt{121}}{2\left(-2\right)}$$

Take the square root of 121.

$$x=\frac{-29±11}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-29±11}{-4}$$

Now solve the equation x=\frac{-29±11}{-4} when ± is plus. Add -29 to 11.

$$x=-\frac{18}{-4}$$

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

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

Now solve the equation x=\frac{-29±11}{-4} when ± is minus. Subtract 11 from -29.

$$x=-\frac{40}{-4}$$

Divide -40 by -4.

$$x=10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{2} for x_{1} and 10 for x_{2}.

$$-2x^{2}+29x-90=-2\left(x-\frac{9}{2}\right)\left(x-10\right)$$

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

$$-2x^{2}+29x-90=-2\times \frac{-2x+9}{-2}\left(x-10\right)$$

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

$$-2x^{2}+29x-90=\left(-2x+9\right)\left(x-10\right)$$

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.

$$x ^ 2 -\frac{29}{2}x +45 = 0$$

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

$$r + s = \frac{29}{2} rs = 45$$

Two numbers r and s sum up to \frac{29}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{29}{2} = \frac{29}{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>

$$r = \frac{29}{4} - u s = \frac{29}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 45

$$(\frac{29}{4} - u) (\frac{29}{4} + u) = 45$$

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

$$\frac{841}{16} - u^2 = 45$$

Simplify the expression by subtracting \frac{841}{16} on both sides

$$-u^2 = 45-\frac{841}{16} = -\frac{121}{16}$$

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

$$u^2 = \frac{121}{16} u = \pm\sqrt{\frac{121}{16}} = \pm \frac{11}{4}$$

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

$$r =\frac{29}{4} - \frac{11}{4} = 4.500 s = \frac{29}{4} + \frac{11}{4} = 10$$

Evaluate

$$\left(10-x\right)\left(2x-9\right)$$