Question

factor each polynomial completely.

3x^2+ 30x+ 75

Answer 1

3x^2+2250x

Answer 2

3(x+5)(x+5)

Answer 3

Hey guys, please explain how you got here to help snsdfb understand the process. Thanks!

Answer 4

each term in the trinomial is factorable by 3, so pull out the 3.

Answer 5

once you do that youll have 3(x^2+10x+25)

Answer 6

so to factor that, find two terms that will multiple to 25, but add to 10, which is 5.

Answer 7

so youll have 3(x+5)(x+5). to check your work, simply multiple the parenthesis and then distribute the 3

Answer 8

thank you can you explain how to factor
5m^2- 20n^2?

Answer 9

both of those terms are divisible by 5. so pull out the 5 from each term. youll have 5(m^2 - 4n^2)

Answer 10

since you have more than one variable you cannot factor any further than that.

Answer 11

okay got it. how about 2k^3-36k^2+162k?

Answer 12

each term is divisible by 2, so divide each term by 2 and youll get 2(k^3-18k^2+81)

Answer 13

x is also a common factor, so take out an x from each term

Answer 14

so youll have 2k(K^2-18k+81)

Answer 15

so now like before find two terms that multiply to positive 81, but add to negative 18, which is -9

Answer 16

so your final answer would be 2k(k-9)(k-9). to check your answer simply multiply the parenthesis then distribute 2k to each term

Answer 17

thank you. 12x^3-20x^2+30x-50?

Answer 18

pull out a 2x from each term: 2x(6x^2+15x-25)

Answer 19

since you have a leading coefficient you can't accurately an answer that has 2 parenthesis. so you will have to plug in each value into the quadratic equation. the quadratic equation is [-b+/-(sqrt((b^2)-(4*a*c)))]/(2*a)

Answer 20

i need to factor the polynomial

Answer 21

once you plug in all the values your answer will be -15+/--5root33 all over 12

Answer 22

that polynomial is not factorable.

Answer 23

okay

Answer 24

factor the polynomial.
6x^2-2x-20

Answer 25

pull out the 2: 2(3x^2-x-10)
using the 3 find two numbers that will multiply to -10 but add to -1: (x-2)(3x+5)

im not here to do your math homework, if youre having trouble with the process of getting the answer just ask me question about that.