How do you multiply (2x²-3x+2)(3x³-x²+√2x)?

2 Answers
Aug 30, 2015

Answer:

Determine the coefficient of each power of #x# in descending order to find:

#(2x^2-3x+2)(3x^3-x^2+sqrt(2)x)#

#=6x^5-11x^4+(9+2sqrt(2))x^3-(2+3sqrt(2))x^2+2sqrt(2)x#

Explanation:

For each power of #x# in descending order, pick out pairs of terms - one from each trinomial - whose product results in that power of #x# and add them together:

#color(green)(x^5)#: #2x^2*3x^3 = color(blue)(6x^5)#

#color(green)(x^4)#: #(2x^2*-x^2) + (-3x*3x^3) = -2x^4-9x^4 = color(blue)(-11x^4)#

#color(green)(x^3)#: #(2x^2*sqrt(2)x) + (-3x*-x^2) + (2*3x^3)#

#= 2sqrt(2)x^3+3x^3+6x^3 = color(blue)((9+2sqrt(2))x^3)#

#color(green)(x^2)#: #(-3x*sqrt(2)x) + (2*-x^2) = -3sqrt(2)x^2-2x^2#

#= color(blue)(-(2+3sqrt(2))x^2)#

#color(gree)(x)#: #2*sqrt(2)x = color(blue)(2sqrt(2)x)#

#color(green)(1)#: #color(blue)(0)#

So:

#(2x^2-3x+2)(3x^3-x^2+sqrt(2)x)#

#=6x^5-11x^4+(9+2sqrt(2))x^3-(2+3sqrt(2))x^2+2sqrt(2)x#

Actually in practice, I would just work with the coefficients:

#color(green)(x^5)#: #2*3 = color(blue)(6)#

#color(green)(x^4)#: #(2*-1) + (-3*3) = -2-9 = color(blue)(-11)#

#color(green)(x^3)#: #(2*sqrt(2)) + (-3*-1) + (2*3) = 2sqrt(2)+3+6 = color(blue)(9+2sqrt(2))#

#color(green)(x^2)#: #(-3*sqrt(2)) + (2*-1) = -3sqrt(2)-2 = color(blue)(-(2+3sqrt(2)))#

#color(green)(x)#: #2*sqrt(2) = color(blue)(2sqrt(2))#

#color(green)(1)#: #color(blue)(0)#

...and I would just write out the sum as I went along:

#6x^5-11x^4+(9+2sqrt(2))x^3-(2+3sqrt(2))x^2+2sqrt(2)x#

Aug 30, 2015

Answer:

#6x^5-11x^4+(9+2sqrt(2))x^3-(2+3sqrt(2))x^2+2sqrt(2)x#

Explanation:

#{: (color(orange)(" X")," | ", color(orange)(2x^2), color(orange)(-3x), color(orange)(+2)), (" -"," | ", "-", "-", "-"), (color(orange)(3x^3)," | ", 6x^5,color(red)( -9x^4), color(blue)(+6x^3)), (color(orange)(-x^2)," | ",color(red)( -2x^4),color(blue)(+3x^3), color(green)(-2x^2)), (color(orange)(+sqrt(2)x)," | ",color(blue)(2sqrt(2)x^3), color(green)(-3sqrt(2)x^2), +2sqrt(2)x) :}#

Combine terms with identical exponents of #x# to get the Answer above.