A Maths Binomial Theorem Qn
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Hi all..Pls help me with this question:
Find in Ascending powers of X, 1st 3 terms of (X^2 -3X + 2)^4
Thx in advance..I rly need it
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Factorise the quadratic equation in the brackets first.
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There is a lazy method:Originally posted by lpx88:Hi all..Pls help me with this question:
Find in Ascending powers of X, 1st 3 terms of (X^2 -3X + 2)^4
Thx in advance..I rly need it
U can convert the above to
( (x-1)(x-2) ) ^ 4
which gives u
( (x-1)^4 )( (x-2)^4 )
in which u can expand easily to
(x^4 - 4x^3 + 6x^2 - 4x + 1) (x^4 - 8x^3 + 24x^2 - 32x + 16)
note that for the first 3 terms, u will get
a x^8 + b x^7 + c x^6 + ...
coefficient of x^8 will thus be the coefficient of the x^4 (from the 1st bracket) multiplied by the coefficient of x^4 (from the 2nd bracket) = 1
similarly, b = (1 * -
+ (-4 * 1) = -12
and c = (1 * 24) + (-4 * - + (6 * 4) = 80
Hence, the first 3 terms are
x^8 - 12x^7 + 80x^6
I may have some careless mistakes in calculations, but the general lazy way is like above...
Hope it helps. -
yes this is the method i did last yrOriginally posted by eagle:There is a lazy method:
U can convert the above to
( (x-1)(x-2) ) ^ 4
which gives u
( (x-1)^4 )( (x-2)^4 )
in which u can expand easily to
A B
(x^4 - 4x^3 + 6x^2 - 4x + 1) (x^4 - 8x^3 + 24x^2 - 32x + 16)
note that for the first 3 terms, u will get
a x^8 + b x^7 + c x^6 + ...
coefficient of x^8 will thus be the coefficient of the x^4 (from the A bracket) multiplied by the coefficient of x^4 (from the B bracket) = 1
similarly, b = (1 * - + (-4 * 1) = -12
and c = (1 * 24) + (-4 * - + (6 * 4) = 80
Hence, the first 3 terms are
x^8 - 12x^7 + 80x^6
I may have some careless mistakes in calculations, but the general lazy way is like above...
Hope it helps. -
(x² -3x + 2)^4Originally posted by lpx88:Hi all..Pls help me with this question:
Find in Ascending powers of x, 1st 3 terms of (x² -3x + 2)^4
Thx in advance..I rly need it
= (x² -3x + 2)(x² -3x + 2)(x² -3x + 2)(x² -3x + 2)
= (2)(2)(2)(2) + (4)(2)(2)(2)(-3x) + (4)(2)(2)(2)(x²) + (6)(2)(2)(-3x)(-3x) + ...
= 16 - 96x + 248x² + ... -
yup... I forgot it was ascending.Originally posted by ^tamago^:(x² -3x + 2)^4
= (x² -3x + 2)(x² -3x + 2)(x² -3x + 2)(x² -3x + 2)
= 16 + (4)(2)(2)(2)(-3x) + (4)(2)(2)(2)(x²) + (4)(2)(2)(-3x)(-3x) + ...
= 16 - 96x + 176x² + ... -
yeah. i forgot term in x² requires 6 permutations on (-3x)(-3x)(2)(2) too... answer updated to 248... it can be derived using eagle's, which i think is more elegant....Originally posted by eagle:yup... I forgot it was ascending. -
yrs is de correct ans as provided by the book but why so many 2's one?Originally posted by ^tamago^:(x² -3x + 2)^4
= (x² -3x + 2)(x² -3x + 2)(x² -3x + 2)(x² -3x + 2)
= (2)(2)(2)(2) + (4)(2)(2)(2)(-3x) + (4)(2)(2)(2)(x²) + (6)(2)(2)(-3x)(-3x) + ...
= 16 - 96x + 248x² + ... -
u take a coefficient each from the 4 quadratics such that they form a term in x^0, x¹ and x².... so for term in x^0 it's 2*2*2*2=16 lor...Originally posted by lpx88:yrs is de correct ans as provided by the book but why so many 2's one? -
1st set is 2*2*2*2..but how come in the subsequent sets got involve 4's and 6's one ahOriginally posted by ^tamago^:u take a coefficient each from the 4 quadratics such that they form a term in x^0, x¹ and x².... so for term in x^0 it's 2*2*2*2=16 lor... -
sry i cant help u now im doing work...but subsequent q's i can help
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use the method I posted... but modify if for ascending powers instead of descending. If u can do it, u will most likely be able to do future questions of a similar natureOriginally posted by lpx88:yrs is de correct ans as provided by the book but why so many 2's one? -
Yea i got it alsoOriginally posted by eagle:use the method I posted... but modify if for ascending powers instead of descending. If u can do it, u will most likely be able to do future questions of a similar nature
So for eg to get coeff of lets say in this case..X^3
so is (1 * 24) + (-4 X -32) + (16 X -6) =248 which is the std ans indeed..
Thx alot
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it's pick and choose lah... dunno how to explain also...Originally posted by lpx88:1st set is 2*2*2*2..but how come in the subsequent sets got involve 4's and 6's one ah 
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Ok..now can someone explain to me about the general term?(R +1)th term?
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It is not really used in this case, because it will be too troublesome. It is usually used when there are only 2 terms in the expansionOriginally posted by lpx88:Ok..now can someone explain to me about the general term?(R +1)th term? -
well done
keep it up -
Woah blur.
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Thx all.I did relatively well for my binomial test 2day
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Don't worry... Just post your questions here.Originally posted by seotiblizzard:Woah blur.
O level and below I should be able to answer all (I hope)
A level I hv forgotten a bit, but should be able to answer at least 70%
Uni level, perhaps first yr physics and maths only...
I'm only an NUS EE undergrad, can still try...