# Math Help - Coefficient of x^3 in (1-3x)(1+2x)^6

1. ## Coefficient of x^3 in (1-3x)(1+2x)^6

how to solve (1-3x)(1+2x)^6?find the coefficient of term x^3 ..

thanks a lot ..

2. Originally Posted by jaycee
how to solve (1-3x)(1+2x)^6?find the coefficient of term x^3 ..

thanks a lot ..
$(1+2x)^6=\sum^{n}_{r=0}(6Cr)(2x)^r$

$(1-3x)(1+2x)^6=\sum^{n}_{r=0}(6Cr)(2x)^r-\sum^{n}_{r=0}(6Cr)2^kx^{k+1}$

so for the first part put r=3 and the second part put r=2.

3. Hello, jaycee!

You could simply crank out the multiplication . . .

$(1-3x)(1+2x)^6$

$\text{Find the coefficient of }x^3$

$(1 + 2x)^6 \;=\;{6\choose6}1 + {6\choose5}(2x) + {6\choose4}(2x)^2 + {6\choose3}(2x)^3 + \hdots$

. . . . . . . . $=\;1 + 12x + 60x^2 + 160x^3 + \hdots$

$(1-3x)(1+2x)^6 \;=\;(1-3x)(1 + 12x + 60x^2 + 160x^3+ + \hdots)$

. . . . . . . . . . . . . $=\; 1 + 12x + 60x^2 + 160x^3 + \hdots$
. . . . . . . . . . . . . . . . . $-\;3x - 36x^2 - 180x^3 - 480x^4 - \hdots$

. . . . . . . . . . . . . $=\;1 + 9x + 24x^2 - 20x^3 + \hdots$

Answer: . $-20$

4. ## thank you

Originally Posted by Soroban
Hello, jaycee!

You could simply crank out the multiplication . . .

$(1 + 2x)^6 \;=\;{6\choose6}1 + {6\choose5}(2x) + {6\choose4}(2x)^2 + {6\choose3}(2x)^3 + \hdots$

. . . . . . . . $=\;1 + 12x + 60x^2 + 160x^3 + \hdots$

$(1-3x)(1+2x)^6 \;=\;(1-3x)(1 + 12x + 60x^2 + 160x^3+ + \hdots)$

. . . . . . . . . . . . . $=\; 1 + 12x + 60x^2 + 160x^3 + \hdots$
. . . . . . . . . . . . . . . . . $-\;3x - 36x^2 - 180x^3 - 480x^4 - \hdots$

. . . . . . . . . . . . . $=\;1 + 9x + 24x^2 - 20x^3 + \hdots$

Answer: . $-20$
thanks alott !

5. thanks !