Discussion in 'The Lounge' started by Leaka, Jan 6, 2008.
Okay now I'm squaring binomials.
I need help again.
(4w+2)^2(exponent of 2)
= (4w+2) (4w+2)
= (16w^2) + (8w) + (8w) + (4)
= 16w^2 + 16w + 4
That's what I believe the answer is... but then again, I suck at math. Sure as hell can't remember how to multiply with larger than binomials...
Adamant got it right Leaka.
A trick for squaring binomials is something like this:
1. Square the first term.
2. Square the second term.
3. Multiply the two together, then multiply that by 2.
1. 4w x 4w = 16w^2
2. 2 x 2 = 4
3. 4w x 2 = 8w x 2 = 16w
Final answer: 16w^2+16w+4
(ax + b)^2 = a^2*x^2 + 2abx + b^2
using w for x, a = 4 and b = 2,
4^2*w^2 + 2(4*2)w + 2^2 = 16w^2 + 16 w + 4
You can easily derive the squaring of a binomial from the product of two binomials:
(ax + b)^2 = (ax + b) * (ax + b) = a^2*x^2[left terms] + axb[outer terms] + bax [inner terms] + b^2 [right terms]
axb and bax are both equivalent to abx , so combining terms you get:
a^&2*x^2 + 2abx + b^2
Cogito, if anything that confused the hell outta me and I already passed this stuff o___o
This isn't the best place to explain this kind of algebra problem, unfortunately.
The best way to understand squaring a binomail is to first understand mutliplying two binomials in the same variable.
(ax + b)(cx +d)
Using the distributive law, this is equivalent to:
ax(cx + d) + b(cx + d)
which is in turn equal to:
acxx + adx + bcx + bd
in terms of the original formula (ax + b)(cx + d):
We call ax*cx as the left product,
b*d the right product
ax*d the outer product
and b*cx the inner product
The result is what we get by adding all 4 of these products, and conventionally order them from the highest order (exponent) of x (x^2) down to the lowest order of x (x^0, which is just a multiplier of 1, which is to say x is not a factor in the product.
Again, the final result is then:
acxx + adx + bcx + bd which is also ac*x^2 + (ad + bc)*x + bd
I hope this makes it a little clearer.
*WOMP WOMP WOMP WOMP WOMP*
So what you're saying is FOIL? Right, FOIL. I like the sound of that.
I don't push mnemonics, because I prefer to help the person understand WHY it works the way it does.
It may be a steeper slope to get there, but it's more level when you arrive.
Cogito, that still confuses me xD
I think that FOIL is easier. Or the way I explained above ^
It may be simpler, but it's only a "how to" formula. What I'm trying to show is why it works, and how to understand the broader case of multiplying two binomials.
If you understand wow to multiply two binomials, you can square a binomial. The steps you give in your post certainly work, but only apply to squaring a binomial.
It's important to understand multiplying binomials; before long you will have to multiply not only binomials, but also polynomials (sums of 3 or more terms). If you are struggling with multiplying binomials, you will be in tears when you have to multiply longer polynomials. That's why I went into the detail I did.
In general, the requirement is:
1. Multiply every term in the first polynomial by every term in the second polynomial, and sum them all together
2. Combine like terms
With two binomials, rule 1 gives you a sum of 4 terms. FOIL works ok for that case. If one of them is a trinomial instead (3 terms), you get a sum of 6 terms, and you can no longer depend on FOIL.
FWIW: I intend to teach mathematics (and science) at the college level in a few years.
THIS IS A WRITING FORUM.
This is where a good book comes handy. Unfortunately, not all Math books are as helpful as others but it would truly benefit you to shop around. I know that in some cases, books just won't do, but with this kind of Algebra, you really shouldn't have a problem deciphering solved examples provided in text books.
You could also look around a library.
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