Finite Math Examples

$9 x - 9$ , $x^{2} - 4 x + 3$ , $x^{2} - 6 x + 9$

Step 1

Factor $9$ out of $9 x - 9$ .

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Step 1.1

Factor $9$ out of $9 x$ .

$9 (x) - 9$

Step 1.2

Factor $9$ out of $- 9$ .

$9 (x) + 9 (- 1)$

Step 1.3

Factor $9$ out of $9 (x) + 9 (- 1)$ .

$9 (x - 1)$

Step 2

Factor $x^{2} - 4 x + 3$ using the AC method.

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Step 2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $- 4$ .

$- 3, - 1$

Step 2.2

Write the factored form using these integers.

$(x - 3) (x - 1)$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rewrite $9$ as $3^{2}$ .

$x^{2} - 6 x + 3^{2}$

Step 3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Step 3.3

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 3 + 3^{2}$

Step 3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

${(x - 3)}^{2}$

Step 4

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 5

$9$ has factors of $3$ and $3$ .

$3 \cdot 3$

Step 6

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 7

The LCM of $9, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$3 \cdot 3$

Step 8

Multiply $3$ by $3$ .

$9$

Step 9

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

Step 10

The factor for $x - 3$ is $x - 3$ itself.

$(x - 3) = x - 3$

$(x - 3)$ occurs $1$ time.

Step 11

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

Step 12

The factors for $x - 3$ are $(x - 3) \cdot (x - 3)$ , which is $x - 3$ multiplied by itself $2$ times.

$(x - 3) = (x - 3) \cdot (x - 3)$

$(x - 3)$ occurs $2$ times.

Step 13

The LCM of $x - 1, x - 3, x - 1, {(x - 3)}^{2}$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x - 1) {(x - 3)}^{2}$

Step 14

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$9 (x - 1) {(x - 3)}^{2}$