QueenV
  • QueenV
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!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
first, factor out b from both terms
QueenV
  • QueenV
Could you show me the work?
anonymous
  • anonymous
how about you show me and I'll correct you if you make any mistakes

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

QueenV
  • QueenV
(b3)x+(49b)x
anonymous
  • anonymous
no... b is the variable instead of x. \[b^{3} +49b= b(b^{2}+49)\]
anonymous
  • anonymous
\[\text{now you can factor the }b^{2} + 49 \text{ term}\]
QueenV
  • QueenV
b*b +7*7
anonymous
  • anonymous
not exactly... what level is your class? have you dealt with imaginary and/or complex numbers?
QueenV
  • QueenV
No, this is Introduction to Algebra. Elementary/Intermediate
anonymous
  • anonymous
by the "ac method" you are talking about factoring, yes?
QueenV
  • QueenV
yes
anonymous
  • 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
  • QueenV
Can you show me what you mean anyway?
anonymous
  • 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
  • anonymous
+ not +
anonymous
  • anonymous
+ not = i mean in the (b=7i) term... it should be b+7i
QueenV
  • QueenV
wow
QueenV
  • 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
  • anonymous
let's focus on the first one... first, write it so the exponents are in descending order (biggest to smallest)
QueenV
  • QueenV
2z+18z+45
anonymous
  • anonymous
i think you mean \[ z^{2} +18z+45\]
QueenV
  • QueenV
was that correct
QueenV
  • QueenV
yes z2
anonymous
  • anonymous
so can you identify what a, b and c are? the coefficients on the z^2, z and constant term, respectively?
QueenV
  • 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
  • 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
  • QueenV
Yes
anonymous
  • 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
  • QueenV
12+6=18, 9+9, 5+13=18, 4+14=18, 2+16=18, 3+15=18
QueenV
  • QueenV
3+15=18, 15*3=45
anonymous
  • 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
  • QueenV
so b is replaced with 15+3?
anonymous
  • anonymous
sorry... \[z^{2}+18z+45 \ldots z(z+15)+3(z+15)\] are the first and last, respectively.
anonymous
  • anonymous
yes... and then the terms are grouped (like above) and factored. can you follow what I did?
QueenV
  • QueenV
Distributive property?
anonymous
  • anonymous
yes... although in reverse from how it is often used...
QueenV
  • QueenV
In reverse?
anonymous
  • anonymous
usually, it's a(b+c) => ab+ac... however we use it in reverse...ab+ac => a(b+c)
anonymous
  • 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
  • QueenV
ok so the next problem is like this one?
anonymous
  • 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
  • QueenV
Common in both terms is a4b and a2b?
anonymous
  • anonymous
yes
QueenV
  • QueenV
(a4b)+(a2)+(b3)?
anonymous
  • 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
  • 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
  • QueenV
Thank you so much pgpilopt326

Looking for something else?

Not the answer you are looking for? Search for more explanations.