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jadewest

Member

Registered: 2021-02-20

Posts: 44

Factoring Polynomials

Hi,
I need help with checking if these are correct.
Thank you!

A.  Factor the following polynomials completely.

1. 125x^3 + 64

(5x)^3 + 43

125x^3 + 64 = (5x + 4) ( (5x)^2 - (5x) (4) + 42 )

(5x + 4) ( 25x^2 - 20x + 16 )

2. x^9 + 1

(x^3)^3 + 1^3
x^9 + 1 = (x + 1) (x^2 - (x) (1) + 1^2)

(x + 1) (x^2 - x + 1)

3. 2m^4 - 2mn^3

2m^3

2m^3 (m) - 2m^3 (n)

2m^3 (m - n)

4. 3a^4 + 81a

3a

3a (a^3) + 3a (27)

3a (a^3 + 27)

B.   Can these be factored using the methods (GCF, squares, cubes) discussed in this lesson? Yes or No?  Explain why or why not in a complete sentence.

5. 7x^5 - 64y

This can't be factored, because 7 and 64 don't share a GCF.

6. a^4 + 1

This can't be factored, because sum of squares cannot be factored using real numbers.

7. a^4 – 64

This can be factored with the difference of squares which leads to the answer: (a^2 + 8) (a^2 - 8)

C.    Factor completely using Grouping

8.  xy – 5y – 2x + 10

y (x - 5) - 2 (x - 5)

(x - 5) (y - 2)

9.  x^3 + x^2 - x – 1

x^2 (x +1) - 1 (x + 1)

(x + 1) (x^2 - 1)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Factoring Polynomials

hi Jade,

Q1.  This is correct. The quadratic cannot be further factored in real numbers.

Q2.  You are right that (x+1) is a factor.  But the other term should contain x^8 etc else how will you get back to x^9 if you expand the bracket.  The other term has x^8, x^7, x^6 ... ... x and a number.  I checked it using a site called wolframalpha which has a super mathematical engine and this long polynomial has two factors.  I don't think you could be expected to get these.

Q3.  This has gone wrong.  The first factor is 2m leaving (m^3-n^3) which can be further factored.

Q4.  Ok so far but (a^3 + 27) can be further factored.

Q5. Correct.

Q6. If you look at your previous post you'll see zetafunc has given you a way to do this one.

Q7.  (a^2-8) can be further factored.

Q8. Correct.

Q9. (x^2-1) can be further factored.

Hope that helps,

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

jadewest

Member

Registered: 2021-02-20

Posts: 44

Re: Factoring Polynomials

Hi,
Do these look correct now?
Thank you!

2. x^9 + 1

(x^3)^3 + 1^3
x^9 + 1 = (x + 1) (x^2 - (x) (1) + 1^2)

(x + 1) (x^2 - x + 1) (x^6 - x^3 + 1)

3. 2m^4 - 2mn^3

2m

2m (m^3) - 2m (m^2 - n^3)

2m (m^3 - m^2 - n^3)

4. 3a^4 + 81a

3a

3a (a^3) + 3a (27)

3a (a^3 + 27)
3a (a^3 + 3^3)

6. a^4 + 1

This can be factored, and leads to the answer (a^2 + 1)^2 -2a^2

7. a^4 – 64

This can be factored with the difference of squares which leads to the answer: (a^2 + 2^3) (a^2 - 2^3)

9.  x^3 + x^2 - x – 1

x^2 (x +1) - 1 (x + 1)

(x + 1) (x^2 - 1)
(x - 1) (x + 1)
(x - 1) (x + 1)^2

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Factoring Polynomials

hi Jade

2. (x + 1) (x^2 - x + 1) (x^6 - x^3 + 1) Yes! Well done.

3. 2m (m^3) - 2m (m^2 - n^3)

Something's gone wrong here. You started with two terms but now you have three when you take out the 2m.

4. 3a^4 + 81a

3a (a^3 + 3^3) the bracket can also be factored.

6. a^4 + 1

This can be factored, and leads to the answer (a^2 + 1)^2 -2a^2 Good.

7. a^4 – 64

This can be factored with the difference of squares which leads to the answer: (a^2 + 2^3) (a^2 - 2^3)

Good, but the second bracket is also the diff of 2 sqs so you can go further.

9.  (x - 1) (x + 1)^2 Good.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

jadewest

Member

Registered: 2021-02-20

Posts: 44

Re: Factoring Polynomials

Hello!

I am not understanding what I am doing wrong in question 3. Also in questions 4 and 7, how do they continue?

Thank you in advance,
Jade

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Factoring Polynomials

hi Jade,

Q3. 2m^4 - 2mn^3

This is 2m.m.m.m - 2m.n.n.n where a dot means multiply

You are right that 2m is a common factor, that leaves

2m(m.m.m - n.n.n)

One way to determine if there are more factors is to notice that the bracket would have value zero if m=n.  That tells me that (m-n) is also a factor

So that gives

2m(m-n)(m.m +m.n + n.n)

Q4. 3a^4 + 81a

3a (a^3 + 27) the bracket can also be factored.

I notice that when a = -3 the bracket is zero so (a+3) is a factor.

3a(a+3)(a.a -3.a +9)

Can this quadratic be factored?

The quadratic formula would have square root(9 - 36) so there are no real roots.  The factorisation is complete.

Q7. a^4 – 64

This can be factored with the difference of squares which leads to the answer: (a^2 + 2^3) (a^2 - 2^3)

I notice that root(2^3) = 2root(2) so the last bracket is also the difference of two squares:

(a^2 + 8)[a + 2root(2)].[a - 2root(2)]

Hope that helps,

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob