Multiplying with fractions
- Thread starter Gr8fu13
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J
JeffM
Guest
Gr8fu13 said:
Well it IS right.
I suspect you do not understand the REASON for the formula. It is not SAFE to use a formula you do not understand because you may use it when you should not. You will also DOUBT yourself, and that is not a good feeling (although one we all become used to).
Let's take a numerical example, 7 * 12. Implicitly, the way you have been taught to do this problem is (7 * 12) = (7 * 2) + (7 * 10), right? You break it down into simpler problems that can be solved with the multiplication table. OK Let's break the more complex expression of [(w + x) * (y + z)] into simpler expressions. Let's do exactly what we did with the 7 and the 12.
[(w + x) * (y + z)] = [w(y + z)] + [x(y + z)]. You understand that step don't you?
Let's use the same method again.
w(y + z) = wy + wz.
x(y + z) = xy + xz. Anything mysterious here?
So, [(w + x) * (y + z)] = [w(y + z)] + [x(y + z)] = wy + wz + xy + xz.
That is the GENERAL result. If you multiply together two expressions each with two summands you get a result with four summands because 2 * 2 = 4.
But wait, what about (a + b) * (a + b).
Well it follows the general rule: (a + b) * (a + b) = (a * a) + (a * b) + (b * a) + (b * b). But that simplifies to
a^2 + ab + ba + b^2. And ab = ba. So we can simplify to
a^2 + ab + ab + b^2. Now we can simplify further to
a^2 + 2ab + b^2.
Now all these letters (w,x,y, z,a, and b) can stand for any number whatsoever. So we have just shown that (a + b) * (a + b) = a^2 + 2ab + b^2 for any pair of numbers a and b. Now you know WHY the formula works and so your "feel" will become more sensitive.
J
JeffM
Guest
Gr8fu13 said:
Well yes and no. All we talked about before was products of two multiplicands, each of which had two summands. Now we are talking about two multiplicands, one with two summands and the other with three. So the answer is not going to follow exactly the same pattern.
BUT the principles involved are the same, and the patterns are consequently somewhat similar.
Did you understand my previous answer, the one that started with 7 * 12?
J
JeffM
Guest
Gr8fu13 said:I did understand your previous example, I just do not get why the 7 is not split up too. I don't understand why we do many of the things in math that we do
OK then I gave a bad example. I have never tried teaching before and am probably exceptionally bad at it.
We multiply 7 * 12 as (7 * 2) + (7 * 10) because we learned our 7 tables way back in 3rd or 4th grade (at least we did back when I was a kid, which was practically the Dark Ages). You already know the answers to (7 * 2) and (7 * 10). So no more simplification was needed. But we COULD HAVE broken the 7 up too.
Please remember the GENERAL rule that we found (u + v)(w + x) = uw + ux + vw + vx. I am going to do (7 * 12) that way. As you know (7 * 12) = 84.
7 = (6 + 1). 12 = (4 + 8). So, (7 * 12) = (6 + 1) * (4 + 8) = (6 * 4) + (6 * 8) + (1 * 4) + (1 * 8) = 24 + 48 + 4 + 8 = 84. We didn't break the problem down that way because it was simpler IN THIS CASE to just break the 12 down into 10 and 2 because you already knew what (7 * 10) and (7 * 2) were.
The absolutely most GENERAL rule in math is to break every hard problem into simpler problems that you already know how to solve.
Now our rule of (u + v)(w + x) = uw + ux + vw + vx is actually a specific example of an even more general rule called the distributive rule, which explains how to multiply any number of sums. I shall show you the pattern right now.
(u + v)(w + x) = uw + ux + vw + vx. The rule we already discussed.
(u + v)(w + x + y) = [u(w + x + y)] + [v(w + x + y)] = uw + ux + uy + vw + vx + vy.
(u + v)(w + x + y + z) = [u(w + x + y + z)] + [v(w + x + y + z)] = uw + ux + uy + uz + vw + vx + vy + vz.
Do you see the pattern?
Now you try one:
(t + u + v)(w + x + y + z) = ?
Once you give me that answer, I'll help you solve your specific problem, but I shall not solve it for you. OK?
J
JeffM
Guest
Gr8fu13 said:
Well it IS right, but you made a mistake in notation.
You forgot the plus signs in the middle part. But I think you have the idea of distribution. If you take it step by step, you will never get confused. Of course there are ways to speed the process up, but they all depend on the basic method. Distribution is the general principle that will let you solve all kinds of problems of this sort. Shall we try your specific question now (the one with the b's) or do you want to try it yourself?
PS I am going to go watch a stupid movie now so I shall not be online again until Sunday. (I watch only stupid movies because the dog gets confused by any other kind.)
JeffM said:
Use this:
(t + u + v)(w + x + y + z) =
t (w + x + y + z) + u( w + x + y + z) + v(w + x + y + z) =
tw + tx + ty + tz + uw + ux + uy + uz + vw + vx + vy + vz
Or this:
\(\displaystyle (t + u + v)(w + x + y + z) =\)
\(\displaystyle t(w + x + y + z) + u(w + x + y + z) + v(w + x + y + z) =\)
\(\displaystyle tw + tx + ty + tz + uw + ux + uy + uz + vw + vx + vy + vz\)
Okay. Let me try doing this for: (b + t)(b^2 - bt t^2) Since there is a subtraction sign this is going to be dealing with negative variables. Correct?
(b + t)(b^2 - bt t^2)
b(b^2 - bt t^2) t(b^2 - bt t^2)
bb^2 - bbt + bt^2 * tb^2 - tbt + tt^2
b^3 - b^2t + bt^2 * tb^2 - t^2b + t^3
bt + bt^2 * t^2b
bt^3 * bt^2
bt^3+2
bt^5 <-----answer?
The directions are to multiply and simplify. Would this be correct?
(b + t)(b^2 - bt t^2)
b(b^2 - bt t^2) t(b^2 - bt t^2)
bb^2 - bbt + bt^2 * tb^2 - tbt + tt^2
b^3 - b^2t + bt^2 * tb^2 - t^2b + t^3
bt + bt^2 * t^2b
bt^3 * bt^2
bt^3+2
bt^5 <-----answer?
The directions are to multiply and simplify. Would this be correct?
J
JeffM
Guest
No. As lookagain has said, you are leaving out plus signs, which may mean that you do not yet understand what you are doing, will probably mean that you make errors in your future computations, and will certainly mean that your teacher will not give you full credit.
The ORIGINAL problem was to simplify (b + t)(b^2 - bt + t^2), right?
When you multiply (7 * 12) using the distributive property you get (7 * 12) = (7 * 10) PLUS (7 * 2) = 70 + 14 = 84. The plus is just as important as the multiplications. You CANNOT forget the plus sign between the two simpler multiplication.
The pattern is:
(u + v)(x + y) = [u(x + y)] + [v(x + y)] = ux + uy + vx + vy
(u + v)(x + y + z) = [u(x + y + z)] + [v(x + y + z)] = ux + uy + uz + vx + vy + vz
(u + v + w)(x + y) = [u(x + y)] + [v(x + y)] + [w(x + y)] = ux + uy + vx + vy + wx + wy
(u + v + w)(x + y + z) = [u(x + y + z)] + [v(x + y + z)] + [w(x + y + z)] = ux + uy + uz + vx + vy + vz + wx +wy + wz
NOTICE THE PLUS SIGN BETWEEN THE SQUARE BRACKETS. To UNDERSTAND you need to grasp the middle step, and why the expressions in square brackets are joined by PLUS signs rather than multiplied. You are adding together the products resulting from multiplying the second factor by each summand in the first factor.
Now you did understand the need to address the minus sign. That was good.
OK Now let's take your work in detail
(b + t)(b^2 - bt + t^2) EQUALS b(b^2 - bt + t^2) PLUS t(b^2 - bt + t^2). It is not just a series of operations but a series of equalities.
The (b + t) gets broken down into b times the second factor and t times the second factor, but the plus sign between b and t does not get lost: it joins the two products. This is the key to the whole process. If you still do not understand, let me know, and I'll ask tkhunny to help you grasp the concept.
You can deal with things in brackets or parentheses without considering what is outside them, right?
Now b(b^2 - btt^2) = (bb^2 - bbt + bt^2).
And t(b^2 - bt + t^2) = (tb^2 - tbt + tt^2).
But now we have to remember that plus sign
So (b + t)(b^2 - bt + t^2) = [b(b^2 - bt + t^2)] + [t(b^2 - bt+ t^2)] = bb^2 - bbt + bt^2 + tb^2 - tbt + tt^2.
Now we simplify by using the rules of exponents.
bb^2 - bbt + bt^2 + tb^2 - tbt + tt^2 = b^3 - tb^2 + bt^2 + tb^2 - bt^2 + t^3 = b^3 + t^3.
You made a mistake because you forgot that plus sign.
I hope this helps.
The ORIGINAL problem was to simplify (b + t)(b^2 - bt + t^2), right?
When you multiply (7 * 12) using the distributive property you get (7 * 12) = (7 * 10) PLUS (7 * 2) = 70 + 14 = 84. The plus is just as important as the multiplications. You CANNOT forget the plus sign between the two simpler multiplication.
The pattern is:
(u + v)(x + y) = [u(x + y)] + [v(x + y)] = ux + uy + vx + vy
(u + v)(x + y + z) = [u(x + y + z)] + [v(x + y + z)] = ux + uy + uz + vx + vy + vz
(u + v + w)(x + y) = [u(x + y)] + [v(x + y)] + [w(x + y)] = ux + uy + vx + vy + wx + wy
(u + v + w)(x + y + z) = [u(x + y + z)] + [v(x + y + z)] + [w(x + y + z)] = ux + uy + uz + vx + vy + vz + wx +wy + wz
NOTICE THE PLUS SIGN BETWEEN THE SQUARE BRACKETS. To UNDERSTAND you need to grasp the middle step, and why the expressions in square brackets are joined by PLUS signs rather than multiplied. You are adding together the products resulting from multiplying the second factor by each summand in the first factor.
Now you did understand the need to address the minus sign. That was good.
OK Now let's take your work in detail
(b + t)(b^2 - bt + t^2) EQUALS b(b^2 - bt + t^2) PLUS t(b^2 - bt + t^2). It is not just a series of operations but a series of equalities.
The (b + t) gets broken down into b times the second factor and t times the second factor, but the plus sign between b and t does not get lost: it joins the two products. This is the key to the whole process. If you still do not understand, let me know, and I'll ask tkhunny to help you grasp the concept.
You can deal with things in brackets or parentheses without considering what is outside them, right?
Now b(b^2 - btt^2) = (bb^2 - bbt + bt^2).
And t(b^2 - bt + t^2) = (tb^2 - tbt + tt^2).
But now we have to remember that plus sign
So (b + t)(b^2 - bt + t^2) = [b(b^2 - bt + t^2)] + [t(b^2 - bt+ t^2)] = bb^2 - bbt + bt^2 + tb^2 - tbt + tt^2.
Now we simplify by using the rules of exponents.
bb^2 - bbt + bt^2 + tb^2 - tbt + tt^2 = b^3 - tb^2 + bt^2 + tb^2 - bt^2 + t^3 = b^3 + t^3.
You made a mistake because you forgot that plus sign.
I hope this helps.
You are multiplying (b^2 - bt + t^2) by (b + t);
so (b^2 - bt + t^2) is multipled by b, then by t (or by t, then by b)
If you can "see" that, you'll be ok :wink:
so (b^2 - bt + t^2) is multipled by b, then by t (or by t, then by b)
Code:
b^2 - bt + t^2
b + t
=================
b^3 - b^2t + bt^2 : (b^2 - bt + t^2) multiplied by b
b^2t - bt^2 + t^3 : (b^2 - bt + t^2) multiplied by t
=======================
b^3 + t^3