And thanks for your happy wishes ðŸ˜€

]]>= 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+âˆš3, and -1-âˆš3.????

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

Er,,,,,,,,

a) f(x)= [x-(-2)][x-(1+âˆš3)][x-(1-âˆš3)]

= (x+2)(x^{2}-2x+1-3)

= x^{3}-6x-4

Ahhh you finally got it right on the `n`^{th} 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 `a``x`^{3}+`b``x`^{2}+`c``x`+`d`, whereby they tried to sub in the root values and solve the resultant simultaneous equations. That's mathematical suicide for you.

Er,,,,,,,,

a) f(x)= (x+2)(x-1)+âˆš3)(x-1)-âˆš3)

= (x+2)(x^2-2x+1-3)

= x^3-6x-2

a) f(x)= (x+2)(x-1)+)âˆš3)(x-1)-âˆš3)

= (x+2)(x-1)^2-âˆš3^2

= (x+2)(x^2-2x+1)-3

= x^3-3x-1 ]]>

Still thinking though ]]>

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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