Find the LCM 8x^2-36x-20 , 2x^2+2x-60

$8 x^{2} - 36 x - 20$ , $2 x^{2} + 2 x - 60$

Step 1

Factor $8 x^{2} - 36 x - 20$ .

Factor $4$ out of $8 x^{2} - 36 x - 20$ .

Factor $4$ out of $8 x^{2}$ .

$4 (2 x^{2}) - 36 x - 20$

Factor $4$ out of $- 36 x$ .

$4 (2 x^{2}) + 4 (- 9 x) - 20$

Factor $4$ out of $- 20$ .

$4 (2 x^{2}) + 4 (- 9 x) + 4 (- 5)$

Factor $4$ out of $4 (2 x^{2}) + 4 (- 9 x)$ .

$4 (2 x^{2} - 9 x) + 4 (- 5)$

Factor $4$ out of $4 (2 x^{2} - 9 x) + 4 (- 5)$ .

$4 (2 x^{2} - 9 x - 5)$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = - 9$ .

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

$4 (2 x^{2} - 9 (x) - 5)$

Rewrite $- 9$ as $1$ plus $- 10$

$4 (2 x^{2} + (1 - 10) x - 5)$

Apply the distributive property.

$4 (2 x^{2} + 1 x - 10 x - 5)$

Multiply $x$ by $1$ .

$4 (2 x^{2} + x - 10 x - 5)$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$4 ((2 x^{2} + x) - 10 x - 5)$

Factor out the greatest common factor (GCF) from each group.

$4 (x (2 x + 1) - 5 (2 x + 1))$

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$4 ((2 x + 1) (x - 5))$

Remove unnecessary parentheses.

$4 (2 x + 1) (x - 5)$

Step 2

Factor $2 x^{2} + 2 x - 60$ .

Factor $2$ out of $2 x^{2} + 2 x - 60$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) + 2 x - 60$

Factor $2$ out of $2 x$ .

$2 (x^{2}) + 2 (x) - 60$

Factor $2$ out of $- 60$ .

$2 (x^{2}) + 2 x + 2 \cdot - 30$

Factor $2$ out of $2 (x^{2}) + 2 x$ .

$2 (x^{2} + x) + 2 \cdot - 30$

Factor $2$ out of $2 (x^{2} + x) + 2 \cdot - 30$ .

$2 (x^{2} + x - 30)$

Factor.

Factor $x^{2} + x - 30$ using the AC method.

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 $- 30$ and whose sum is $1$ .

$- 5, 6$

Write the factored form using these integers.

$2 ((x - 5) (x + 6))$

Remove unnecessary parentheses.

$2 (x - 5) (x + 6)$

Step 3

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 4

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

Step 5

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 6

The LCM of $4, 2$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2 \cdot 2$

Step 7

Multiply $2$ by $2$ .

$4$

Step 8

The factor for $2 x + 1$ is $2 x + 1$ itself.

$(2 x + 1) = 2 x + 1$

$(2 x + 1)$ occurs $1$ time.

Step 9

The factor for $x - 5$ is $x - 5$ itself.

$(x - 5) = x - 5$

$(x - 5)$ occurs $1$ time.

Step 10

The factor for $x + 6$ is $x + 6$ itself.

$(x + 6) = x + 6$

$(x + 6)$ occurs $1$ time.

Step 11

The LCM of $2 x + 1, x - 5, x - 5, x + 6$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(2 x + 1) (x - 5) (x + 6)$

Step 12

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

$4 (2 x + 1) (x - 5) (x + 6)$

Find the LCM 8x^2-36x-20 , 2x^2+2x-60

Related Questions

Support LCMGCF.com by disabling your adblocker.