What is the standard form of #y(64y + 1)(y + 25) #?

Standard form of a polynomial means to write it like this:

#a*y^n + b*y^(n-1) + c*y^(n-2) + cdots + p * y + q#

Where the terms of the polynomial are written in order of decreasing exponents.

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In this case, let's start by expanding the two terms #(64y+1)(y+25)#. We can use the FOIL method to do so:

#"FIRST"#

#(color(red)(64y)+1)(color(red)y+25) => color(red)(64y * y) = color(red)(64y^2#

#"OUTER"#

#(color(blue)(64y)+1)(y+color(blue)25) => color(blue)(64y * 25) = color(blue)(1600y#

#"INNER"#

#(64y+color(limegreen)1)(color(limegreen)y+25) => color(limegreen)(1*y) = color(limegreen)(y#

#"LAST"#

#(64y + color(orange)1)(y+color(orange)25) => color(orange)(1*25) = color(orange)(25#

So our polynomial is:

#(64y+1)(y+25) = color(red)(64y^2) + color(blue)(1600y) + color(limegreen)y + color(orange)25 = 64y^2 + 1601y + 25#

Finally, remember that all of this was multiplied by #y# in the original expression:

#y(64y+1)(y+25)#

So, we need to multiply our polynomial by #color(orange)y# to get the final standard form of the polynomial:

#color(orange)y(64y^2+1601y+25) = 64y^2*color(orange)y + 1601y*color(orange)y + 25 * color(orange)y#

#= 64y^3 + 1601 y^2 + 25y#

Final Answer