How do you write #y= (x+3) (x+4) (x+5)# in standard form?

1 Answer
Apr 11, 2016

Answer:

#y = x^3+12x^2+47x+60#

Explanation:

The quickest way to multiply this out is probably to look at each power of #x# in descending order from #3# to #0#, pick out the choices of terms from each binomial factor that combine to give the power of #x# and add them together:

#x^3# : The only way you can get a term in #x^3# is by multiplying all of the leading terms of the binomials together: #x xx x xx x#. So the coefficient of #x^3# is #color(blue)(1)#.

#x^2# : The way that you can get a term in #x^2# is by choosing one of the binomials to provide a constant term and picking the #x# term from the other binomials. Hence the coefficient of #x^2# in the product of the three binomials is #3+4+5 = color(blue)(12)#.

#x# : The way that you can get a term in #x# is by choosing one of the binomials to provide the #x# and the other two to provide constant multipliers. Hence the coefficient of #x# in the product of the three binomials is #4*5+5*3+3*4 = 20+15+12 = color(blue)(47)#.

#1# : The constant term in the product contains no factor of #x#, so the only way you can get it is by multiplying all of the three constants from the binomials together: #3*4*5 = color(blue)(60)#

Hence we find:

#y = (x+3)(x+4)(x+5) = x^3+12x^2+47x+60#

I have described the process in a fairly lengthy way, but with practice, you can probably do these sums mostly in your head, gathering together the various combinations of terms.