Factor the polynomial by grouping x^3-3x^2+5x-15?

Answer 1

x^2(x-3)+5(x-3)
(x^2+5)(x-3)

Answer 2

really you do this... the first difficulty says (5x2 + 5)(x + a million) what the parentises recommend is multiplycation. in addition they recommend the numbers interior the parentises bypass first. hence the numbers interior the paretises bypass toghther. you may't take a quantity from the different set of parentises. the x continually = a million. because a million x some thing continues to be the same answer. like 7, is the same as a million, circumstances 7. so x circumstances 7 will be like a million circumstances 7. anyhow 5x2 is like declaring 5 x a million x 2 which = 10. thus 10 x, becauyse we dont no the authentic cost of x, its basically replaced with the help of a million, because a million x some thing is continually the same. so we've 10x. so as i reported you make certain the to parentises set of numbers in instruments. first one set then the different. now theres plus 5. so that you presently have 10x + 5. you cant make it 15x (10+5, becayse theres so x after 5. so its (10x + 5) now we've were given x + a million that's like declaring a million + a million yet considering we do not no the cost of x (x ought to = some thing) it basically turns into (a million + x) because in case you made it 2 (a million + x) the answer would come out diverse. because lets say x = 5. 5 x 2 = 10. at the same time as 5 x a million continues to be 5. get it? it continues to be the same. so weve now were given (10x + 5) + (1x) 10 is the same as 5 x 2. hence A is the superb perfect selection for quantity one. do some thing else the same way. desire this facilitates!(=-

Answer 3

First: when you have 4 terms > you group "like" terms:

(x^3 - 3x^2) + (5x - 15)

Second: factor each set > find the least common factor:

x^2(x - 3) + 5(x - 3)

third: take one inner term and multiply by the outer term:

(x - 3)(x^2 + 5)

Answer 4

Factoring:
(x³-3x²)+(5x-15) =
x²(x - 3) + 5(x - 3) =
(x² + 5)(x - 3)
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Answer 5

Notice that the ratio of exponents of the first 2 terms (1:-3) is the same as the ratio of exponents of the second 2 terms (5:-15).

So, you can factor out an x^2 from the first 2 terms, and a 5 from the second two, and you've got:

x^2(x - 3) + 5(x - 3)

then you can factor out an (x-3) from both terms:

(x - 3)(x^2 + 5)

And that's as far as you can go (unless you're allowed to use complex numbers, in which case it would be (x - 3)(x + i sqrt(5))(x - i sqrt(5))).

Answer 6

From the first two terms factor out x^2 and from the last two terms, factor out a 5

x^2 (x-3) + 5 (x-3)

Factor out (x-3) from each of the terms above to get

(x-3) (x^2+5)

The sum of squares cannot be factored, so we are done.

Answer 7

x^3 - 3x^2 + 5x - 15
(x^3 - 3x^2) + (5x - 15)
x^2(x - 3) + 5(x - 3)
(x^2 + 5)(x - 3)

Answer 8

x^3-3x^2+5x-15?
= x^2(x-3) + 5(x-3)
=(x^2+5)(x-3)

Answer 9

(x^3-3x^2)+(5x-15)

factor out the x^2 in the (x^3-3x^2)

x^2(x-3)+(5x-15)

factor the 5 out of the (5x-15)

x^2(x-3)+5(x-3)

combine outside terms

(x^2+5)(x-3)

Answer 10

x^3-3x^2+5x-15=
x^2(x-3)+5(x-3)........(x-3) shows twice, so you only chose it once
so..then you group it
(x^2+5)(x-3) is the final answer .. .