Solve 100x^2+60x+9

$$100 { x }^{ 2 } +60x+9$$

Factor

$$\left(10x+3\right)^{2}$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 100x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=60 ab=100\times 9=900$$

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 900.

$$1,900 2,450 3,300 4,225 5,180 6,150 9,100 10,90 12,75 15,60 18,50 20,45 25,36 30,30$$

Calculate the sum for each pair.

$$1+900=901 2+450=452 3+300=303 4+225=229 5+180=185 6+150=156 9+100=109 10+90=100 12+75=87 15+60=75 18+50=68 20+45=65 25+36=61 30+30=60$$

The solution is the pair that gives sum 60.

$$a=30 b=30$$

Rewrite 100x^{2}+60x+9 as \left(100x^{2}+30x\right)+\left(30x+9\right).

$$\left(100x^{2}+30x\right)+\left(30x+9\right)$$

Factor out 10x in the first and 3 in the second group.

$$10x\left(10x+3\right)+3\left(10x+3\right)$$

Factor out common term 10x+3 by using distributive property.

$$\left(10x+3\right)\left(10x+3\right)$$

Rewrite as a binomial square.

$$\left(10x+3\right)^{2}$$

Steps Using Square Of Binomial

This trinomial has the form of a trinomial square, perhaps multiplied by a common factor. Trinomial squares can be factored by finding the square roots of the leading and trailing terms.

$$factor(100x^{2}+60x+9)$$

Find the greatest common factor of the coefficients.

$$gcf(100,60,9)=1$$

Find the square root of the leading term, 100x^{2}.

$$\sqrt{100x^{2}}=10x$$

Find the square root of the trailing term, 9.

$$\sqrt{9}=3$$

The trinomial square is the square of the binomial that is the sum or difference of the square roots of the leading and trailing terms, with the sign determined by the sign of the middle term of the trinomial square.

$$\left(10x+3\right)^{2}$$

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.

$$100x^{2}+60x+9=0$$

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.

$$x=\frac{-60±\sqrt{60^{2}-4\times 100\times 9}}{2\times 100}$$

Square 60.

$$x=\frac{-60±\sqrt{3600-4\times 100\times 9}}{2\times 100}$$

Multiply -4 times 100.

$$x=\frac{-60±\sqrt{3600-400\times 9}}{2\times 100}$$

Multiply -400 times 9.

$$x=\frac{-60±\sqrt{3600-3600}}{2\times 100}$$

Add 3600 to -3600.

$$x=\frac{-60±\sqrt{0}}{2\times 100}$$

Take the square root of 0.

$$x=\frac{-60±0}{2\times 100}$$

Multiply 2 times 100.

$$x=\frac{-60±0}{200}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{10} for x_{1} and -\frac{3}{10} for x_{2}.

$$100x^{2}+60x+9=100\left(x-\left(-\frac{3}{10}\right)\right)\left(x-\left(-\frac{3}{10}\right)\right)$$

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

$$100x^{2}+60x+9=100\left(x+\frac{3}{10}\right)\left(x+\frac{3}{10}\right)$$

Add \frac{3}{10} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$100x^{2}+60x+9=100\times \frac{10x+3}{10}\left(x+\frac{3}{10}\right)$$

Add \frac{3}{10} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$100x^{2}+60x+9=100\times \frac{10x+3}{10}\times \frac{10x+3}{10}$$

Multiply \frac{10x+3}{10} times \frac{10x+3}{10} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$100x^{2}+60x+9=100\times \frac{\left(10x+3\right)\left(10x+3\right)}{10\times 10}$$

Multiply 10 times 10.

$$100x^{2}+60x+9=100\times \frac{\left(10x+3\right)\left(10x+3\right)}{100}$$

Cancel out 100, the greatest common factor in 100 and 100.

$$100x^{2}+60x+9=\left(10x+3\right)\left(10x+3\right)$$

Evaluate

$$\left(10x+3\right)^{2}$$