Question

I have to do this problem by using the ac method, ax2+bx+c. The problem is b3+49b. I have no earthly idea of what to do. Help!

Answer 1

first, factor out b from both terms

Answer 2

Could you show me the work?

Answer 3

how about you show me and I'll correct you if you make any mistakes

Answer 4

(b3)x+(49b)x

Answer 5

no... b is the variable instead of x.
\[b^{3} +49b= b(b^{2}+49)\]

Answer 6

\[\text{now you can factor the }b^{2} + 49 \text{ term}\]

Answer 7

b*b +7*7

Answer 8

not exactly... what level is your class? have you dealt with imaginary and/or complex numbers?

Answer 9

No, this is Introduction to Algebra. Elementary/Intermediate

Answer 10

by the "ac method" you are talking about factoring, yes?

Answer 11

yes

Answer 12

well, if you haven't dealt with complex or imaginary numbers, then you probably should leave it as is...
\[ b^{3}+49b=b(b^{2}+49)\]

Answer 13

Can you show me what you mean anyway?

Answer 14

\[ i =\sqrt{-1} \,\text{ is an imaginary number}\]
\[ b^{2}+49 = (b+7i)(b-7i) \text{ would be the complete factorization of this term and}\]
\[ b^{3}+49b = b(b=7i)(b-7i) \text{ would be the cmplete factorization.}\]

Answer 15

+ not +

Answer 16

+ not = i mean in the (b=7i) term...

it should be b+7i

Answer 17

wow

Answer 18

I have 2 more problems I need help with. If you choose you can give an example. Polynomial (Factor) 18z+45+z2 and Polynomial (Factor Completely) a4b+a2b3. Could the foil method be used for both?

Answer 19

let's focus on the first one...
first, write it so the exponents are in descending order (biggest to smallest)

Answer 20

2z+18z+45

Answer 21

i think you mean
\[ z^{2} +18z+45\]

Answer 22

was that correct

Answer 23

yes z2

Answer 24

so can you identify what a, b and c are? the coefficients on the z^2, z and constant term, respectively?

Answer 25

I am 40 going back to college for Behavioral Science. No excuse, but I have no early idea. I am looking at examples that look like greek to me. I need a tutor. I havent taken algebra in years. please be patient with me. I need step by step.

Answer 26

no worries...
ax^2 +bx +c in your case you have z^2 +18z+45, so z is the variable instead of x. but no matter... we want the numbers associated with the z^2 term... that is a.

in your problem, there is no number written but it is understood to be 1. this is because 1*z^2 = z^2. Make sense? so a = 1.

the number associated with the z term is 18. that one should be fairly obvious. so b = 18.

the constant term is the term with no variable associated with it. it is 45. so c = 45.

do you follow this?

Answer 27

Yes

Answer 28

so the ac method works like this...

multiply a times c to get the product.

the factors of the product must sum to the value of b.

look at this...

Answer 29

12+6=18, 9+9, 5+13=18, 4+14=18, 2+16=18, 3+15=18

Answer 30

3+15=18, 15*3=45

Answer 31

excellent!
so here's what we do with this...

\[z{[2} +18z+45=z^{2} +15z+3z+45=(z^{2}+15z)+(3z+45)=z(z+15)+3(z+15)\]

so we have (z+15) as a common term and we can factor that out...

\[ z(z+15)+3(z+15)=(z+15)(z+3)\]

and now we have factored the polynomial...

\[z^{2}+18z+45 = (z+15)(z+3)\]

Answer 32

so b is replaced with 15+3?

Answer 33

sorry...
\[z^{2}+18z+45 \ldots z(z+15)+3(z+15)\] are the first and last, respectively.

Answer 34

yes... and then the terms are grouped (like above) and factored. can you follow what I did?

Answer 35

Distributive property?

Answer 36

yes... although in reverse from how it is often used...

Answer 37

In reverse?

Answer 38

usually, it's a(b+c) => ab+ac... however we use it in reverse...ab+ac => a(b+c)

Answer 39

the property is a(b+c) = ab+ac but most only think to use this in going from the left to the right... that's why i said in reverse because we're going from the right to the left.

Answer 40

ok so the next problem is like this one?

Answer 41

similar... you always want to factor out anything that is common to your terms.
you have
\[a^{4}b+a^{2}b^{3}\] so what is common to both terms? factor that out (distributive property in action again)...

Answer 42

Common in both terms is a4b and a2b?

Answer 43

yes

Answer 44

(a4b)+(a2)+(b3)?

Answer 45

not like that...

\[ a^{2}b \text{ is common to both. So I can factor that out of both terms...}\]
\[a^{4}b+a^{2}b^{3} =a^{2}b(a^{2}+b^{2})\]

Answer 46

@DebbieG can you help? I'm sorry QueenV but I have to leave now. Hopefully DebbieG can help you.
Have a great day!

Answer 47

Thank you so much pgpilopt326