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

Jul 3, 2017

$64 {y}^{3} + 1601 {y}^{2} + 25 y$

#### Explanation:

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

$a \cdot {y}^{n} + b \cdot {y}^{n - 1} + c \cdot {y}^{n - 2} + \cdots + p \cdot y + q$

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 $\left(64 y + 1\right) \left(y + 25\right)$. We can use the FOIL method to do so:

$\text{FIRST}$

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

$\text{OUTER}$

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

$\text{INNER}$

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

$\text{LAST}$

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

So our polynomial is:

$\left(64 y + 1\right) \left(y + 25\right) = \textcolor{red}{64 {y}^{2}} + \textcolor{b l u e}{1600 y} + \textcolor{\lim e g r e e n}{y} + \textcolor{\mathmr{and} a n \ge}{25} = 64 {y}^{2} + 1601 y + 25$

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

$y \left(64 y + 1\right) \left(y + 25\right)$

So, we need to multiply our polynomial by $\textcolor{\mathmr{and} a n \ge}{y}$ to get the final standard form of the polynomial:

$\textcolor{\mathmr{and} a n \ge}{y} \left(64 {y}^{2} + 1601 y + 25\right) = 64 {y}^{2} \cdot \textcolor{\mathmr{and} a n \ge}{y} + 1601 y \cdot \textcolor{\mathmr{and} a n \ge}{y} + 25 \cdot \textcolor{\mathmr{and} a n \ge}{y}$

$= 64 {y}^{3} + 1601 {y}^{2} + 25 y$