Question

Factor completely and then place the factors in the proper location on the grid.

Answer 1

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 2

Answer:

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

$mn+3m-5n-15$

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

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