Solve 10b^2-9b-1

$$10 b ^ { 2 } - 9 b - 1$$

Factor

$$\left(b-1\right)\left(10b+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10b^{2}+pb+qb-1. To find p and q, set up a system to be solved.

$$p+q=-9 pq=10\left(-1\right)=-10$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -9.

$$p=-10 q=1$$

Rewrite 10b^{2}-9b-1 as \left(10b^{2}-10b\right)+\left(b-1\right).

$$\left(10b^{2}-10b\right)+\left(b-1\right)$$

Factor out 10b in 10b^{2}-10b.

$$10b\left(b-1\right)+b-1$$

Factor out common term b-1 by using distributive property.

$$\left(b-1\right)\left(10b+1\right)$$

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.

$$10b^{2}-9b-1=0$$

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.

$$b=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 10\left(-1\right)}}{2\times 10}$$

Square -9.

$$b=\frac{-\left(-9\right)±\sqrt{81-4\times 10\left(-1\right)}}{2\times 10}$$

Multiply -4 times 10.

$$b=\frac{-\left(-9\right)±\sqrt{81-40\left(-1\right)}}{2\times 10}$$

Multiply -40 times -1.

$$b=\frac{-\left(-9\right)±\sqrt{81+40}}{2\times 10}$$

Add 81 to 40.

$$b=\frac{-\left(-9\right)±\sqrt{121}}{2\times 10}$$

Take the square root of 121.

$$b=\frac{-\left(-9\right)±11}{2\times 10}$$

The opposite of -9 is 9.

$$b=\frac{9±11}{2\times 10}$$

Multiply 2 times 10.

$$b=\frac{9±11}{20}$$

Now solve the equation b=\frac{9±11}{20} when ± is plus. Add 9 to 11.

$$b=\frac{20}{20}$$

Divide 20 by 20.

$$b=1$$

Now solve the equation b=\frac{9±11}{20} when ± is minus. Subtract 11 from 9.

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

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

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

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

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

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

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

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

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

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

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

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

$$x ^ 2 -\frac{9}{10}x -\frac{1}{10} = 0$$

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

$$r + s = \frac{9}{10} rs = -\frac{1}{10}$$

Two numbers r and s sum up to \frac{9}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{10} = \frac{9}{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>

$$r = \frac{9}{20} - u s = \frac{9}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{10}

$$(\frac{9}{20} - u) (\frac{9}{20} + u) = -\frac{1}{10}$$

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

$$\frac{81}{400} - u^2 = -\frac{1}{10}$$

Simplify the expression by subtracting \frac{81}{400} on both sides

$$-u^2 = -\frac{1}{10}-\frac{81}{400} = -\frac{121}{400}$$

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

$$u^2 = \frac{121}{400} u = \pm\sqrt{\frac{121}{400}} = \pm \frac{11}{20}$$

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

$$r =\frac{9}{20} - \frac{11}{20} = -0.100 s = \frac{9}{20} + \frac{11}{20} = 1$$

Evaluate

$$\left(b-1\right)\left(10b+1\right)$$