Solve -x^2+140x-1300

$$- x ^ { 2 } + 140 x - 1300$$

Factor

$$\left(10-x\right)\left(x-130\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-1300. To find a and b, set up a system to be solved.

$$a+b=140 ab=-\left(-1300\right)=1300$$

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 1300.

$$1,1300 2,650 4,325 5,260 10,130 13,100 20,65 25,52 26,50$$

Calculate the sum for each pair.

$$1+1300=1301 2+650=652 4+325=329 5+260=265 10+130=140 13+100=113 20+65=85 25+52=77 26+50=76$$

The solution is the pair that gives sum 140.

$$a=130 b=10$$

Rewrite -x^{2}+140x-1300 as \left(-x^{2}+130x\right)+\left(10x-1300\right).

$$\left(-x^{2}+130x\right)+\left(10x-1300\right)$$

Factor out -x in the first and 10 in the second group.

$$-x\left(x-130\right)+10\left(x-130\right)$$

Factor out common term x-130 by using distributive property.

$$\left(x-130\right)\left(-x+10\right)$$

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.

$$-x^{2}+140x-1300=0$$

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.

$$x=\frac{-140±\sqrt{140^{2}-4\left(-1\right)\left(-1300\right)}}{2\left(-1\right)}$$

Square 140.

$$x=\frac{-140±\sqrt{19600-4\left(-1\right)\left(-1300\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-140±\sqrt{19600+4\left(-1300\right)}}{2\left(-1\right)}$$

Multiply 4 times -1300.

$$x=\frac{-140±\sqrt{19600-5200}}{2\left(-1\right)}$$

Add 19600 to -5200.

$$x=\frac{-140±\sqrt{14400}}{2\left(-1\right)}$$

Take the square root of 14400.

$$x=\frac{-140±120}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-140±120}{-2}$$

Now solve the equation x=\frac{-140±120}{-2} when ± is plus. Add -140 to 120.

$$x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x=10$$

Now solve the equation x=\frac{-140±120}{-2} when ± is minus. Subtract 120 from -140.

$$x=-\frac{260}{-2}$$

Divide -260 by -2.

$$x=130$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and 130 for x_{2}.

$$-x^{2}+140x-1300=-\left(x-10\right)\left(x-130\right)$$

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.

$$x ^ 2 -140x +1300 = 0$$

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

$$r + s = 140 rs = 1300$$

Two numbers r and s sum up to 140 exactly when the average of the two numbers is \frac{1}{2}*140 = 70. 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>

$$r = 70 - u s = 70 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1300

$$(70 - u) (70 + u) = 1300$$

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

$$4900 - u^2 = 1300$$

Simplify the expression by subtracting 4900 on both sides

$$-u^2 = 1300-4900 = -3600$$

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

$$u^2 = 3600 u = \pm\sqrt{3600} = \pm 60$$

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

$$r =70 - 60 = 10 s = 70 + 60 = 130$$

Evaluate

$$\left(10-x\right)\left(x-130\right)$$