These are easy if you know " FOIL ". That's a procedure that takes you through multiplying two binomials.
FOIL stands for
-- <u><em>F</em></u>irst terms
<span>-- </span><u><em>O</em></u>utside terms
<span>--<em> </em></span><u><em>I</em></u>nside terms
<span>--<em> </em><u><em>L</em></u></span>ast terms
and that's how you keep everything straight while you're doing it.
(ax + b) x (cx + d)
Multiply First terms . . . 'ax' times 'cx' = <em>acx²</em>
Multiply Outside terms . . . 'ax' times 'd' = <em>adx</em>
Multiply Inside terms . . . 'b' times 'cx' = <em>bcx</em>
Multiply Last terms . . . 'b' times 'd' = <em>bd</em>
Now addummup:
(ax + b) x (cx + d) = <em>acx² + adx + bcx + bd</em>
From there, you can look for opportunities to make it look cleaner and prettier ... factoring, combining like terms, etc.
Please look at the picture for a step by step explanation - Answer choice is A
Answer:

Step-by-step explanation:
Given,
Length of a rectangular garden ( l ) = 2y - 4
Width of a rectangular garden ( w ) = 4y²
Perimeter of the garden = ?
<u>Finding</u><u> </u><u>the</u><u> </u><u>perimeter</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>rectangular</u><u> </u><u>garden</u><u> </u><u>having </u><u>length </u><u>of</u><u> </u><u>2</u><u>y</u><u>-</u><u>4</u><u> </u><u>and</u><u> </u><u>width</u><u> </u><u>of</u><u> </u><u>4</u><u>y</u><u>²</u>


Distribute 2 through the parentheses

Arrange it in standard form

Hope I helped!
Best regards! :D
Answer:x + y = 54
x - y = 10
Add them to get: 2x = 64
x = 32
and plugging in x = 32 to either equation: y = 22
Ans: the numbers are 22 and 32
Step-by-step explanation:
<u>Here are your fill-ins:</u>
r = 1 is a zero, so (r-1) is a factor.
r = -1 is a zero, so (r+1) is a factor.
The remainders from synthetic division are 0 each time.
See the attachment for the synthetic division numbers.
_____
The reduced quadratic is ...
... x² -6x +13 = 0
Solving by completing the square, we have ...
... (x -3)² = -4
... x = 3 ± 2i
_____
The quadratic formula would tell you ...
... x = (-(-6) ± √((-6)² -4(1)(13)))/(2·1) = (6±√-16)/2 = 3±2i