Solve 10x^2-89x-9

$$10 x ^ { 2 } - 89 x - 9$$

Factor

$$\left(x-9\right)\left(10x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=-89 ab=10\left(-9\right)=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -89.

$$a=-90 b=1$$

Rewrite 10x^{2}-89x-9 as \left(10x^{2}-90x\right)+\left(x-9\right).

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

Factor out 10x in 10x^{2}-90x.

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

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(10x+1\right)$$

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.

$$10x^{2}-89x-9=0$$

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.

$$x=\frac{-\left(-89\right)±\sqrt{\left(-89\right)^{2}-4\times 10\left(-9\right)}}{2\times 10}$$

Square -89.

$$x=\frac{-\left(-89\right)±\sqrt{7921-4\times 10\left(-9\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-89\right)±\sqrt{7921-40\left(-9\right)}}{2\times 10}$$

Multiply -40 times -9.

$$x=\frac{-\left(-89\right)±\sqrt{7921+360}}{2\times 10}$$

Add 7921 to 360.

$$x=\frac{-\left(-89\right)±\sqrt{8281}}{2\times 10}$$

Take the square root of 8281.

$$x=\frac{-\left(-89\right)±91}{2\times 10}$$

The opposite of -89 is 89.

$$x=\frac{89±91}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{89±91}{20}$$

Now solve the equation x=\frac{89±91}{20} when ± is plus. Add 89 to 91.

$$x=\frac{180}{20}$$

Divide 180 by 20.

$$x=9$$

Now solve the equation x=\frac{89±91}{20} when ± is minus. Subtract 91 from 89.

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

Reduce the fraction \frac{-2}{20} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{10}$$

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

$$10x^{2}-89x-9=10\left(x-9\right)\left(x-\left(-\frac{1}{10}\right)\right)$$

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

$$10x^{2}-89x-9=10\left(x-9\right)\left(x+\frac{1}{10}\right)$$

Add \frac{1}{10} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}-89x-9=10\left(x-9\right)\times \frac{10x+1}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10x^{2}-89x-9=\left(x-9\right)\left(10x+1\right)$$

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 10

$$x ^ 2 -\frac{89}{10}x -\frac{9}{10} = 0$$

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

$$r + s = \frac{89}{10} rs = -\frac{9}{10}$$

Two numbers r and s sum up to \frac{89}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{89}{10} = \frac{89}{20}. 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>

$$r = \frac{89}{20} - u s = \frac{89}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{10}

$$(\frac{89}{20} - u) (\frac{89}{20} + u) = -\frac{9}{10}$$

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

$$\frac{7921}{400} - u^2 = -\frac{9}{10}$$

Simplify the expression by subtracting \frac{7921}{400} on both sides

$$-u^2 = -\frac{9}{10}-\frac{7921}{400} = -\frac{8281}{400}$$

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

$$u^2 = \frac{8281}{400} u = \pm\sqrt{\frac{8281}{400}} = \pm \frac{91}{20}$$

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

$$r =\frac{89}{20} - \frac{91}{20} = -0.100 s = \frac{89}{20} + \frac{91}{20} = 9$$

Evaluate

$$\left(x-9\right)\left(10x+1\right)$$