Solve 9x^2+10x+1

$$9 { x }^{ 2 } +10x+1$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=10 ab=9\times 1=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=1 b=9$$

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

$$\left(9x^{2}+x\right)+\left(9x+1\right)$$

Factor out x in 9x^{2}+x.

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

Factor out common term 9x+1 by using distributive property.

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

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.

$$9x^{2}+10x+1=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 9}}{2\times 9}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 9}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-10±\sqrt{100-36}}{2\times 9}$$

Add 100 to -36.

$$x=\frac{-10±\sqrt{64}}{2\times 9}$$

Take the square root of 64.

$$x=\frac{-10±8}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-10±8}{18}$$

Now solve the equation x=\frac{-10±8}{18} when ± is plus. Add -10 to 8.

$$x=-\frac{2}{18}$$

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

$$x=-\frac{1}{9}$$

Now solve the equation x=\frac{-10±8}{18} when ± is minus. Subtract 8 from -10.

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

Divide -18 by 18.

$$x=-1$$

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

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

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

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

Add \frac{1}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 9, the greatest common factor in 9 and 9.

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

Evaluate

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