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!

- QueenV

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- anonymous

first, factor out b from both terms

- QueenV

Could you show me the work?

- anonymous

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

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## More answers

- QueenV

(b3)x+(49b)x

- anonymous

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

- anonymous

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

- QueenV

b*b +7*7

- anonymous

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

- QueenV

No, this is Introduction to Algebra. Elementary/Intermediate

- anonymous

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

- QueenV

yes

- anonymous

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

- QueenV

Can you show me what you mean anyway?

- anonymous

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

- anonymous

+ not +

- anonymous

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

- QueenV

wow

- QueenV

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?

- anonymous

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

- QueenV

2z+18z+45

- anonymous

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

- QueenV

was that correct

- QueenV

yes z2

- anonymous

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

- QueenV

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.

- anonymous

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?

- QueenV

Yes

- anonymous

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...|dw:1378420385080:dw|

- QueenV

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

- QueenV

3+15=18, 15*3=45

- anonymous

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

- QueenV

so b is replaced with 15+3?

- anonymous

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

- anonymous

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

- QueenV

Distributive property?

- anonymous

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

- QueenV

In reverse?

- anonymous

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

- anonymous

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.

- QueenV

ok so the next problem is like this one?

- anonymous

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)...

- QueenV

Common in both terms is a4b and a2b?

- anonymous

yes

- QueenV

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

- anonymous

not like that...
\[ a^{2}b \text{ is common to both. So I can factor that out of both terms...}\]
\[a^{4}b+a^{2}b^{3} =a^{2}b(a^{2}+b^{2})\]

- anonymous

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

- QueenV

Thank you so much pgpilopt326

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