3U MATHS on Binomial Help! (1 Viewer)

littleremedy · Jul 20, 2012

Hi ! Can you please help me work this out. Its based on ''greatest co-efficients'' in Binomials

Thankyou

barbernator · Jul 20, 2012

by simple inspection, the coefficient of x^3 will be (-5)(-3)^3(4C1)

you can write a formula for these questions to find out what term will include the x^3 term, yet in this expansion is is obvious that is is the term (-5x)(x^2)(-3)^3(4C1)

littleremedy · Jul 20, 2012

RealiseNothing · Jul 20, 2012

Barbernator already answered your question.....

Aykay · Jul 20, 2012

$x^{3}\ term\ in\ (2-5x)(x^{2}-3)^{4} \\ \\ Expanding:\ (2-5x)\left \{ \binom{4}{0}(x^{2})^{4}\ -\ \binom{4}{1}(x^{2})^{3}(3)^{1}\ +\ \binom{4}{2}(x^{2})^{2}(3)^{2}\ -\ \binom{4}{3}(x^{2})^{1}(3)^{3}\ +\ \binom{4}{4}(3)^{4}\ \right \} \\ \\ So\ we're\ looking\ for\ an\ x^{3}\ or\ x^{2}\ term\ in\ the\ binomial\ expansion\ to\\ \\ multiply\ by\ (2-5x)\ so\ we\ can\ get\ an\ x^{3}\ term. \\ \\ There's\ only\ an\ x^{2}\ term,\ so\ we\ multiply\ it\ with\ the\ x^{1}\ term\ in\ (2-5x)\\ \\ \therefore\ x^{3}\ term\ is\ -5x \times -\binom{4}{3}(x^{2})^{1}(3)^{3} \\ \\ =\ 540x^{3}$

First time using latex sorry.

3U MATHS on Binomial Help! (1 Viewer)

littleremedy

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barbernator

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RealiseNothing

what is that?It is Cowpea

Aykay

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