Mathematics, 28.07.2019 02:00 AlmightyThadd

Factor completely and then place the factors in the proper location on the grid. 8c 2 - 26c + 15 find the greatest common factor of 270 and 360. (give the answer in the numerical form in the top box and in exponential form by filling in the boxes for exponents.) [] 2[]3[]5[] (these are where the exponents go[]) find the greatest common factor of 8 a3b2 and 12 ab4 . [] a[]b[] factor completely and then place the factors in the proper location on the grid. 81a 2 + 36a + 4 factor the four-term polynomial. mn - 15 + 3 m - 5 n (m + 3)(n - 5) (n - 3)(m - 5) (n + 3)(m - 5)

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1. 8c^2-26c+15= (4c-3) (2c-5). Break the expression into groups: =(8c^2-6c)+(-20c+15). Factor out 8c^2-6c: 2c(4c-3). Factor out -5 from -20c+ 15: -5(4c-3). Lastly factor out common term (4c-3) and thats how you'll get your answer (4c-3) (2c-5).

2. common factors for 270 and 360 is 90.To find this write the factors of each and find the largest one.270: 1, 270, 2, 135, 3, 90, 5, 54, 6, 45, 9, 30, 10, 27, 15, 18360: 1, 360, 2, 180, 3, 120, 4, 90, 5, 72, 6, 60, 8, 45, 9, 40, 10, 36, 12, 30, 15, 24, 18, 20
3. The factors for 8 a3b2 and 12 ab4 is 4. because 8: 1, 2, 4, 812: 1, 2, 3, 4, 6, 12.

4. 81a^2+36a+4= (9a+2)^2. Break down the expression into groups: (81a^2+18a)+(18a+4). Factor out 9a from 81a^2 +18a: 9a(9a+2). Factor out 2 from 18a+4: 2(9a+2). so the groups you got are now 9a(9a+2)+2(9a+2). Lastley factor out common term (9a+2) to get (9a+2) (9a+2). Finally you get the answer (9a+2)^2.

5. mn-15+3m-5n= (n+3)(m-5). factor out m from nm+3m: m(n+3). Factor out -5 from -5n-15: -5(n+3). And thats how you get the number (n+3)(m-5)

Hope this helped :) Have a great day

Answer from: eemorales5100

1. (2c-5)(4c-3)

2. $2\times 3^2\times 5$

3. 4ab^2

4. (9a+2)(9a+2)

4. (m-5)(n+3)

Step-by-step explanation:

1.We are given that

8c^2-26c+15

We have to find the factor

Using factorization method

8c^2-20c-6c+15

4c(2c-5)-3(2c-5)

(2c-5)(4c-3)

2.We are given that 270 and 360

We have to find the HCF of 270 and 360

$270=2\times 3^3\times 5$

$360=2^3\times3^2\times 5$

Hence, HCF of 270 and 360 = $2\times 3^2\times 5$

3.We are given that 8a^3b^2 and 12ab^4

We have to find HCF

$8a^3b^2=2^3\times a\times a\times a\times b\times b$

$12 ab^4=2^2\times 3\times a\times b\times b\times b\times b$

Common factors are 4,a, b^2

Hence,HCF 8a^3b^2 and 12 ab^4 =4ab^2

4. 81a^2+36a+4

We have to find the factors

81a^2+18a+18a+4

9a(9a+2)+2(9a+2)

(9a+2)(9a+2)

5. mn-15+3m-5n

We have to find the factors

mn+3m-5n-15

m(n+3)-5(n+3)

(m-5)(n+3)

Answer from: Quest

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

Answer from: Quest

relating to or denoting a system of numbers and arithmetic based on the number ten, tenth parts, and powers of ten.

it alot will look like this 4.4

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Factor completely and then place the factors in the proper location on the grid. 8c 2 - 26c + 15 fin...