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

1 Answer
Jul 3, 2017

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

Explanation:

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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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