Multiplying with Polynomials.

Question

In $(3xy)^2$, do I distribute that power of two to each of the terms?

$(3^2)\times(x^2)\times(y^2) = 9x^2y^2$?

Or do I just treat it as $3xy^2$?

$\begingroup$ The first option. $\endgroup$ – Edward Jiang Dec 19 '14 at 2:28 — Edward Jiang, Dec 19 '14 at 2:28
$\begingroup$ $ 3^2 \cdot x^2 \cdot y^2 \; = \; 9x^2y^2 $ $\endgroup$ – mathamphetamines Dec 19 '14 at 2:29 — mathamphetamines, Dec 19 '14 at 2:29
$\begingroup$ you have computer so search it up. on google $\endgroup$ – Irrational Person Dec 19 '14 at 2:31 — Irrational Person, Dec 19 '14 at 2:31
$\begingroup$ I am searching it up, right here. ._. $\endgroup$ – John Dec 19 '14 at 2:32 — John, Dec 19 '14 at 2:32

JMoravitz · Accepted Answer · 2014-12-19 02:29:56Z

6

$$(3xy)^2 = (3xy)(3xy) = 3\cdot 3\cdot x\cdot x\cdot y\cdot y = 3^2x^2y^2=9x^2y^2$$

answered Dec 19 '14 at 2:29

JMoravitz

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$\begingroup$ you got it first $\endgroup$ – Irrational Person Dec 19 '14 at 2:32

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turkeyhundt · Accepted Answer · 2014-12-19 02:30:12Z

1

All terms. $(3xy)^2=(3xy)(3xy)=(9x^2y^2)$

Good time to note how it's important to put those parentheses around expressions like this to distinguish $(3xy)^2$ from $3xy^2$

answered Dec 19 '14 at 2:30

turkeyhundt

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