Help with an algebra factoring problem.

In summary, the conversation discusses a problem involving factoring and expanding expressions. The person asks for help on what to do once they reach a certain form, and are given guidance to factor out 5(x-3) and then expand and combine like terms to reach the final answer. They express gratitude for the explanation and mention their previous confusion.
  • #1
gohabsgo
12
0
Sorry to ask help on such an easy problem. I'm self teaching, but for some reason I'm getting stuck on a step. Can anyone explain to me what to do once I reach the form below, prior to the answer?

1) 15(x-3)3 + 60x4(x-3)2 + 5(x-3)




I get this far, as shown below. After this I'm confused as to what to do next. For instance, (x-3)^2, could be x^2-9, allowing me to use difference of squares, but I'm at a loss when I look at the answer my book gives. Somewhere I'm forgetting something. As for 12x^4, I understand that it's been distributed to give the 12x^5-36x^4, but after that I'm completely lost.
3) 5(x-3) [3(x-3)2 + 12x4(x-3) +1]

According to my material the Answer is: 5(x-3)[12x5 - 36x4 +3x2 -18x +28]
 
Physics news on Phys.org
  • #2
just expand the second part and you are okay
 
  • #3
gohabsgo said:
Sorry to ask help on such an easy problem. I'm self teaching, but for some reason I'm getting stuck on a step. Can anyone explain to me what to do once I reach the form below, prior to the answer?

1) 15(x-3)3 + 60x4(x-3)2 + 5(x-3)




I get this far, as shown below. After this I'm confused as to what to do next. For instance, (x-3)^2, could be x^2-9, allowing me to use difference of squares
No, (x - 3)2 = x2 - 6x + 9, which is different from x2 - 9.
gohabsgo said:
, but I'm at a loss when I look at the answer my book gives. Somewhere I'm forgetting something. As for 12x^4, I understand that it's been distributed to give the 12x^5-36x^4, but after that I'm completely lost.
Factor 5(x - 3) from each of the three terms above, to get what you have just below here. Then expand everything inside the brackets and combine like terms.
gohabsgo said:
3) 5(x-3) [3(x-3)2 + 12x4(x-3) +1]

According to my material the Answer is: 5(x-3)[12x5 - 36x4 +3x2 -18x +28]
 
  • #4
Mark44 said:
No, (x - 3)2 = x2 - 6x + 9, which is different from x2 - 9.
Factor 5(x - 3) from each of the three terms above, to get what you have just below here. Then expand everything inside the brackets and combine like terms.

Oh wow Mark44, thanks. So all I had to do was distribute 3 through (x2 -6x +9) and combine like-terms? I think where I became confused was with the lone 3. I didn't think to keep x2 -6x +9 in brackets to thereafter distribute 3. It's all so clear now and makes perfect sense. Thanks again for the help on something I was literally going crazy over. :)
 

What is factoring in algebra?

Factoring in algebra is the process of breaking down a polynomial expression into smaller, simpler expressions. This is done by finding the common factors that can be divided out of the original expression.

Why is factoring important in algebra?

Factoring is important in algebra because it helps simplify complex expressions, making them easier to solve. It also helps identify patterns and relationships between different algebraic expressions.

How do you factor a polynomial expression?

To factor a polynomial expression, you can use different methods such as finding common factors, grouping terms, or using the difference of squares or sum/difference of cubes formulas. It is important to follow the correct steps and use the appropriate method for each expression.

What is the difference between factoring and simplifying?

Factoring and simplifying are two different processes in algebra. Factoring involves breaking down a polynomial expression into smaller terms, while simplifying involves reducing an expression to its simplest form by combining like terms and performing operations. Factoring is a crucial step in simplifying an expression, but they are not the same thing.

Can factoring be used to solve equations?

Yes, factoring can be used to solve equations. When solving equations, factoring can help identify the roots or solutions of the equation. It is an important tool in solving quadratic equations and can also be used in solving higher degree equations.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
19
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
10
Views
940
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
708
  • Precalculus Mathematics Homework Help
Replies
8
Views
775
  • Precalculus Mathematics Homework Help
Replies
7
Views
887
  • Precalculus Mathematics Homework Help
Replies
10
Views
880
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
Back
Top