How do you multiply #(x-7)(5x^2-3x-5)#?

Question

Hazel Anglin · Accepted Answer

#5x^3-38x^2+16x+35# #"each term in the second bracket is multiplied by each "# #"term in the first bracket, as follows."# #(color(red)(x-7))(5x^2-3x-5)# #=color(red)(x)(5x^2-3x-5)color(red)(-7)(5x^2-3x-5)# #"distributing each bracket gives"# #=5x^3-3x^2-5x-35x^2+21x+35# #"collect like terms for simplification"# #=5x^3-38x^2+16x+35#

Avery Alexander · Answer

#color(green)(5x^3-38x^2+16x+35#

#(x-7)(5x^2-3x-5)#

#:.x xx 5x^2=5x^3#

#:.x xx -3x=-3x^2#

#:.x xx -5=-5x#

#:.-7 xx 5x^2=-35x^2#

#:.-7 xx -3x=21x#

#:.-7 xx -5=35#

#:.=5x^3-3x^2-35x^2-5x+21x+35#

#:.color(green)(=5x^3-38x^2+16x+35#

or:-

#color(white)(aaaaaaaaaaaaa)##5x^2-3x-5# #color(white)(aaaaaaaaaaa)## xx underline(x-7)# #color(white)(aaaaaaaaaaaaa)##5x^3-3x^2-5x# #color(white)(aaaaaaaaaaaaaaaa)##-35x^2+21x+35# #color(white)(aaaaaaaaaaaaa)##overline(5x^3-38x^2+16x+35)#

#color(white)(aaaaaaaaaaaaa)##color(green)(5x^3-38x^2+16x+35#

Ethan Adkins · Answer

undefined To multiply $(x-7)(5x^2-3x-5)$, we use the distributive property:

$$ (x-7)(5x^2-3x-5) = x(5x^2-3x-5) - 7(5x^2-3x-5) $$

$$ = 5x^3 - 3x^2 - 5x - 35x^2 + 21x + 35 $$

$$ = 5x^3 - 38x^2 + 16x + 35 $$

Therefore, $(x-7)(5x^2-3x-5) = 5x^3 - 38x^2 + 16x + 35$.