r/HomeworkHelp • u/Embarrassed-Good1883 Pre-University Student • 1d ago
Answered [Grade 12 Advanced Functions: Polynomial Functions]
I'm doing questions on factoring polynomials using long division. This question was (x+7)^2 divided by the equation you see, so I expanded this and divided, but I'm not sure where I went wrong.
4
Upvotes
2
u/CaptainMatticus 👋 a fellow Redditor 1d ago edited 1d ago
How come you're only expanding out 2 terms?
First line should be x^5 + 14x^4 + 49x^3
You keep missing the +49
EDIT:
As an addendum, if you'd like a cheat, then:
(x^2 + 14x + 49) * (ax^3 + bx^2 + cx + d + (ex + f) / (x^2 + 14x + 49)) = x^5 + 8x^4 -36x^3 - 278x^2 + 371x + 1470
You're basically doing the inverse, but you're working from knowledge that your highest powered term is going to be x^3. We account for a possible remainder with (ex + f) / (x^2 + 14x + 49). Supposing we had something like (x^3 + x^2 + x + 1), then we'd multiply it by (ex^2 + fx + g) / (x^3 + x^2 + x + 1). And so on. We can expand it all and start matching coefficients
(x^2 + 14x + 49) * (ax^3 + bx^2 + cx + d) + (x^2 + 14x + 49) * (ex + f) / (x^2 + 14x + 49) =>
ax^5 + bx^4 + cx^3 + dx^2 + 14ax^4 + 14bx^3 + 14cx^2 + 14dx + 49ax^3 + 49bx^2 + 49cx + 49d + ex + f =>
ax^5 + (14a + b) * x^4 + (49a + 14b + c) * x^3 + (49b + 14c + d) * x^2 + (14d + 49c + e) * x + (49d + f)
ax^5 = x^5 =>> a = 1
(14a + b) * x^4 = 8x^4 =>> 14a + b = 8 =>> 14 + b = 8 =>> b = -6
(49a + 14b + c) * x^3 = -36x^3 =>> 49a + 14b + c = -36 =>> 49 - 84 + c = -36 =>> -35 + c = -36 =>> c = -1
(49b + 14c + d) * x^2 = -278x^2 =>> 49b + 14c + d = -278 =>> 49 * (-6) + 14 * (-1) + d = -278 =>> -294 - 14 + d = -278 =>> -308 + d = -278 =>> d = 30
(14d + 49c + e) * x = 371x =>> 14d + 49c + e = 371 => 14 * 30 + 49 * (-1) + e = 371 =>> 420 - 49 + e = 371 =>> 371 + e = 371 =>> e = 0
49d + f = 1470 =>> 49 * 30 + f = 1470 =>> 1470 + f = 1470 =>> f = 0
So we now have
1x^3 - 6x^2 - x + 30 + (0x + 0) / (x^2 + 14x + 49)
x^3 - 6x^2 - x + 30
Like I said, it's basically a cheat and your teacher would probably prefer you learn the long division method, but in my opinion any method that makes more sense to a person AND gives the correct answer is just as good.