Comments on: Celebrating 100 Posts of Blog-Building With Polynomials http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1417 Miss Loi Wed, 10 Oct 2007 03:06:40 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1417 123, Miss Loi has marked your answers. Next time try getting it right on your first attempt, like what you should do in your actual exams. And thanks for your happy wishes :D 123, Miss Loi has marked your answers. Next time try getting it right on your first attempt, like what you should do in your actual exams.

And thanks for your happy wishes :D

]]> By: kiroii http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1409 kiroii Tue, 09 Oct 2007 15:09:01 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1409 hmm should be correct i got de same ans also hmm should be correct i got de same ans also

]]> By: 123 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1406 123 Tue, 09 Oct 2007 11:25:58 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1406 Ops, i forgot, Happpy 100th post...hope there will be more 100ths to come ^.^ Ops, i forgot, Happpy 100th post...hope there will be more 100ths to come ^.^

]]> By: 123 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1405 123 Tue, 09 Oct 2007 11:25:05 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1405 f(3)=3<sup>3</sup>-6(3)-4 = 5 <div class="highlight"> Yes Part 2 tests your understanding of the <strong>Remainder Theorem</strong> i.e. <strong>If f(<var>x</var>) is divided by (<var>x</var>-<var>a</var>), the remainder is f(<var>a</var>)</strong>. Unfortunately, many Last-Minute Buddha Foot Huggers who quickly flipped through the pages of their textbooks missed this. And proceeded to waste their time in the exam doing <em>long division</em>! And do note that Part 2 requires your answer from Part 1 to be correct in order for it to be correct. So this is a <em>one die all die</em> question. And you should pay extra attention to these kind of questions when you're checking for careless mistakes during your exam. </div> Just making a lucky guess for part (iii), is the answer 2, -1+√3, and -1-√3.???? <div class="highlight"> Yes you're lucky. Some day, if you're taking A-Level Math, you'll be taught that f(-<var>x</var>) is actually a reflection of f(<var>x</var>) about the <var>y</var>-axis. But for now, some students will get stuck in this coz they've never seen it in their textbooks. Instead of just sitting there and staring blankly at the question, simply sub in -<var>x</var> into the original function to get f(-<var>x</var>), i.e. f(<var>x</var>) = (<var>x</var>+2)[<var>x</var>-(1+√3)][<var>x</var>-(1-√3)] = 0 f(-<var>x</var>) = [(-<var>x</var>)+2][(-<var>x</var>)-(1+√3)][(-<var>x</var>)-(1-√3)] = 0 (-<var>x</var>+2)(-<var>x</var>-1-√3)(-<var>x</var>-1+√3)=0 (<var>x</var>-2)[<var>x</var>-(-1-√3)][<var>x</var>-(-1+√3)]=0 ⇒ <var>x</var>=2, <var>x</var>=-1-√3, <var>x</var>=-1+√3 </div> f(3)=33-6(3)-4
= 5

Yes Part 2 tests your understanding of the Remainder Theorem i.e. If f(x) is divided by (x-a), the remainder is f(a).

Unfortunately, many Last-Minute Buddha Foot Huggers who quickly flipped through the pages of their textbooks missed this. And proceeded to waste their time in the exam doing long division!

And do note that Part 2 requires your answer from Part 1 to be correct in order for it to be correct. So this is a one die all die question. And you should pay extra attention to these kind of questions when you're checking for careless mistakes during your exam.

Just making a lucky guess for part (iii), is the answer 2, -1+√3, and -1-√3.????

Yes you're lucky. Some day, if you're taking A-Level Math, you'll be taught that f(-x) is actually a reflection of f(x) about the y-axis.

But for now, some students will get stuck in this coz they've never seen it in their textbooks.

Instead of just sitting there and staring blankly at the question, simply sub in -x into the original function to get f(-x), i.e.

f(x) = (x+2)[x-(1+√3)][x-(1-√3)] = 0
f(-x) = [(-x)+2][(-x)-(1+√3)][(-x)-(1-√3)] = 0
(-x+2)(-x-1-√3)(-x-1+√3)=0
(x-2)[x-(-1-√3)][x-(-1+√3)]=0

⇒ x=2, x=-1-√3, x=-1+√3

]]> By: 123 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1404 123 Tue, 09 Oct 2007 11:22:01 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1404 OOOOOOOOOOOOOppppps wrong again -.-!! Er,,,,,,,, a) f(x)= [x-(-2)][x-(1+√3)][x-(1-√3)] = (x+2)(x<sup>2</sup>-2x+1-3) = x<sup>3</sup>-6x-4 <div class="highlight"> Ahhh you finally got it right on the <var>n</var><sup>th</sup> attempt. Have edited your workings for better readability. When you see the keyword <em>roots</em>, you'll know that Part 1 tests your understanding of the <strong>Factor Theorem</strong>: <strong><var>x</var>-<var>a</var> is a factor of f(<var>x</var>) where f(<var>a</var>) = 0. And <var>x</var>=<var>a</var> is a root of the equation</strong>. Which you've rightly applied. Unfortunately some students got too excited and instead formed an arbitrary <var>a</var><var>x</var><sup>3</sup>+<var>b</var><var>x</var><sup>2</sup>+<var>c</var><var>x</var>+<var>d</var>, whereby they tried to sub in the root values and solve the resultant simultaneous equations. That's mathematical suicide for you. </div> OOOOOOOOOOOOOppppps wrong again -.-!!

Er,,,,,,,,
a) f(x)= [x-(-2)][x-(1+√3)][x-(1-√3)]
= (x+2)(x2-2x+1-3)
= x3-6x-4

Ahhh you finally got it right on the nth attempt. Have edited your workings for better readability.

When you see the keyword roots, you'll know that Part 1 tests your understanding of the Factor Theorem: x-a is a factor of f(x) where f(a) = 0. And x=a is a root of the equation. Which you've rightly applied.

Unfortunately some students got too excited and instead formed an arbitrary ax3+bx2+cx+d, whereby they tried to sub in the root values and solve the resultant simultaneous equations. That's mathematical suicide for you.

]]> By: 123 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1403 123 Tue, 09 Oct 2007 11:20:48 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1403 OOps wrong -.- Er,,,,,,,, a) f(x)= (x+2)(x-1)+√3)(x-1)-√3) = (x+2)(x^2-2x+1-3) = x^3-6x-2 OOps wrong -.-

Er,,,,,,,,
a) f(x)= (x+2)(x-1)+√3)(x-1)-√3)
= (x+2)(x^2-2x+1-3)
= x^3-6x-2

]]> By: 123 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1402 123 Tue, 09 Oct 2007 11:17:07 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1402 Er,,,,,,,, a) f(x)= (x+2)(x-1)+)√3)(x-1)-√3) = (x+2)(x-1)^2-√3^2 = (x+2)(x^2-2x+1)-3 = x^3-3x-1 Er,,,,,,,,
a) f(x)= (x+2)(x-1)+)√3)(x-1)-√3)
= (x+2)(x-1)^2-√3^2
= (x+2)(x^2-2x+1)-3
= x^3-3x-1

]]> By: 123 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1401 123 Tue, 09 Oct 2007 10:56:23 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1401 Haha, thnx but.... this question seems abit the complicated @.@ Still thinking though Haha, thnx but.... this question seems abit the complicated @.@
Still thinking though

]]> By: kiroii http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1399 kiroii Tue, 09 Oct 2007 09:44:36 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1399 hmm this seems easy nvm i'll let 123 solve it since he succinctly lamented he din get to solve de last one hmm this seems easy nvm i'll let 123 solve it since he succinctly lamented he din get to solve de last one

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1397 Miss Loi Tue, 09 Oct 2007 09:22:53 +0000 http://www.exampaper.com.sg/questions/a-maths/celebrating-100-posts-of-blog-building-with-polynomials#comment-1397 HAM, <em>you</em> don't mind but wouldn't your gf mind you dining around with other girls in S'pore??? Now that this 100th post milestone had quietly come and gone (sadly without the promised fanfare), time to think of another grand vision for this blog's anniversary. Maybe we'll see a fly-past of aeroplanes by then! HAM, you don't mind but wouldn't your gf mind you dining around with other girls in S'pore???

Now that this 100th post milestone had quietly come and gone (sadly without the promised fanfare), time to think of another grand vision for this blog's anniversary. Maybe we'll see a fly-past of aeroplanes by then!

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