Please Help ill give medal Rewrite x^2 6x+9 as a square of a linear expression.
douc bag
This is a blog where people help if u don't want to help and make fun of me then stay out of my way.
I just need a more clear cut information o how to solve this that's all.
You're not doing yourself any favors by consistently posting questions without showing any effort. This site exists to help you learn, not to do your homework for you. Without asking specific questions, or at the minimum showing your attempt, you do not provide any information to potential helpers regarding what you are struggling with. In fact, it is a sign of general laziness, and thus, you have a fantastic career at McDonald's awaiting you. Put effort into your work and it will pay off.
Well as you should know math isn't my strongest subject but I have As and Bs in pretty much everything else plus I can work FOR my dad.
Just because math isn't your strong subject doesn't mean you can be lazy with it. Math is involved with everything and it will improve your ability to analyze problems and develop a clearer perspective of seemingly unrelated situations. So, I ask again, what's your attempt, and what are you struggling to understand?
I just don't understand how to rewrite this problem.
Have you covered factoring in class?
Hint: you need to find two numbers that multiply to get 9, and add to get 6. Both numbers will be positive, since you have all positive signs in your expression.
I.e., all terms in your expression are positive.
A linear expression is an expression of degree 1.
like 2nd degree
Your expression is a 2nd degree polynomial, yes. The problem states to turn it into the square of a linear expression, i.e., the product of two linear expressions (degree 1) that are the same. You are trying to get this expression into the form \[(a+b)^2\]
Which is the same as: \[(a+b)(a+b)\]
yes I figured that the (a+b)^2
Alright, so what is your a, and what is your b? What two numbers multiply to get 9 and add to get 6?
my a is 6 and b is 9
\[(a + b)(a+b) = (a^2 + ab + ba + b^2)\] No, \[(a+b)^2\] is the FORM of your final expression. Again, what two numbers multiply to get 9 and add to get 6? This is not a difficult question. Answer it.
For example, 4 and 4 multiply to get 16, and add to get 8.
oh 3*3 and 3+3
Yes, correct. So, your two numbers are 3 and 3 (they happen to be the same for this problem). They are your "b" terms in the form \[(a+b)(a+b)\] Now what is your "a" term?
im not really sure actually.
It's x. Try this: \[(x+3)(x+3) = x(x+3) + 3(x+3)\] right?
oh... I guess that makes sense
If you distribute the x across the first (x+3) and the 3 across the 2nd (x+3) and combine like terms, you end up with your original equation, i.e., the "un-factored" equation. So you know you did it correctly.
ok what is i.e.
i.e. is short-hand for "that is" e.g. is short-hand for "for example" You can remember the difference by thinking "egg-zample"
Remember that \[(a+b)(a+b)\]is just the FORM of the factored expression. You are trying to get this expression into that FORM. The "b's" in that form are the two numbers that MULTIPLY to get your LAST term of the "un-factored" expression as well as ADD to get the MIDDLE term of the "un-factored" equation.
So its (x+3)(x+3)?
The form can and will change depending on your original expression. If you had a negative sign in the middle and a positive at the end then it would be in the form (a-b)(a-b) because two negative multiply to get a positive, and add to get a negative. Yes, (x+3)(x+3) is your answer.
Oh well then I should have told u it was a -6 I honestly didn't even see it
In the future, please initially show an attempt at your work or at least ask specific questions (or explain what you struggle with).
Well, if it is a -6 then you know what to do now, right?
its (x-3)^2 correct?
Good job. :)
ok well before u leave im going to need to make sure the next part is correct.
The part is Do the expression in Parts A and B have a common factor? Justify your answer. The answer I got for A is (x-3) and (x-5) What is the question asking?
Is this a different problem?
So would they be the same because of the x
no
It just has like different parts like Part A and b and c
we did B
Well, imagine this. Imagine that you have two expressions and you place one over the other in a fraction. \[\frac{(a+b)(c+d)}{(a+b)}\] You now that "anything over itself is 1," right? That is, if I had 45, and divided it by 45, the answer would be 1. Because of that, you know that you can "cancel" the common factors in the numerator and denominator of the fraction, right?
yes
So, the question is basically asking, what are the common factors among the factored expression from part A, and the factored expression from part B?
But u see Part A asked for two answers (x-3) and (x-5)
Well I imagine that you factored an expression to find those factors, right? Was the original expression in Part A: \[x^2 -8x + 15\] ?
Part A: Write the expression x^2-18x+45 as a product of two linear expressions. Show you work and justify each step.
Alright, well you did Part A slightly wrong. What two numbers multiply to get 45 and add to get -18?
Hint: both numbers will be negative, since they add to get a negative number, and multiply to get a positive number.
the (x+3) was wrong wasn't it?
5*9 and -8+-10
Yes, but they have to be the same two numbers for both multiplication and addition.
So no, those are not the right numbers.
5 * 9 comes to thought first because we memorized our multiplication tables. This multiplication is not normally memorized in elementary school. 3 * 15 = 45. 3 + 15 = 18. Right? But you need negative numbers, so what are they?
22.5*2 -9+-9
Yes, but they have to be the same two numbers for both multiplication and addition. Key: SAME NUMBERS.
For both addition and multiplication.... 3 * 4 = 12 and 3 + 4 = 7. Those are two numbers that multiply to get 12 and add to get 7.
Notice that they are the same numbers used for both addition and multiplication....
oh so -9+-9 and 9*5
-3 * -15 = 45 -3 + -15 = -18 Same numbers used for both mult. and addition.
oh
Anyway, so your factored expression for Part A is (x-3)(x-15), right? So what is the common factor among the factored expressions from Part A and Part B?
Think back to the example I gave involving "canceling" terms in a fraction.
(x-3)
right?
I don't see your answer.
(x-3) is the same factor
Yes, that's right. Now that your done, I would like you to, in your own words, explain to me how to factor this equation: \[x^2 + 5x + 6\]
can u change your 5x
Nope, that's the expression as it stands. What is your first step? How will you approach this problem?
"What two numbers..."
that 5 is kind of confusing me
I'll state it one more time. Commit this to memory. You need to find two numbers that when multiplied together get 6, and when added together get 5.
3+2 3*2
Right. Good. So, you understand that when looking at x2+5x+6 you need to find two numbers that multiply to get 6 and add to get 5, right?
correct
That's your first step. Now, you need to look at the signs. All the terms are positive. What does this suggest about your factored form?
that its posotive
Right, that all factors have positive terms.
So, what is the factored form of \[x^2+5x+6\] ?
(x-2)
No minus signs. You just said that all terms in the factored form are positive, right?
You are getting the form (a+b)(a+c) here, where a and c are your two numbers.
oh whoops (x+2) (x+3)
Yes, good job. So thanks for sticking around and showing some effort. I hope you have a better understanding of how to approach these problem. Now, for the analysis, note that this is for simple quadratic expression in the form x^2 + bx + c, e.g., x^2 + 5x + 6. If you start seeing more complicated expressions like 27x^2 + 5x + 99 (notice you have a coefficient, 27, on your leading term), the process is a little more complicated. But finding "two numbers that multiply to get... the last term, and add to get the coefficient of the middle term" is a good start. And looking at the signs.
Well if that does happen ill come looking for you to get some help on that but for now I have to got to bed but I have one more question for math that I will need help on from you tomorrow. Will that be fine?
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