The polynomial remainder theorem states that the remainder upon dividing a polynomial

by

is the same as the value of

, so to find

you need to find the remainder upon dividing

You have
..... | 2 ... 14 ... -58
-10 | ... -20 ... 60
--------------------------
..... | 2 ... -6 .... 2
So the quotient and remainder upon dividing is

with a remainder of 2, which means

.
All of Randy’s thing in his desk are round


- <u>We </u><u>have </u><u>given </u><u>that </u><u>the </u><u>coordinates </u><u>of </u><u>the </u><u>end </u><u>point </u><u>G </u><u>and </u><u>H </u><u>are </u><u>(</u><u> </u><u>-</u><u>6</u><u>,</u><u>5</u><u>)</u><u> </u><u>and </u><u>(</u><u> </u><u>2</u><u>,</u><u> </u><u>-</u><u>7</u><u> </u><u>)</u>

- <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>length </u><u>of </u><u>GH </u>

The coordinates of G = ( -6 , 5 )
The coordinates of H = ( 2 , - 7 )
<u>According </u><u>to </u><u>the </u><u>distance </u><u>formula</u><u>, </u><u> </u><u>we </u><u>get </u><u>:</u><u>-</u><u> </u>

- <u>Here</u><u>, </u><u> </u><u>x1</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>,</u><u> </u><u>x2</u><u> </u><u>=</u><u> </u><u>2</u><u> </u><u>and </u><u>y1</u><u> </u><u>=</u><u> </u><u>5</u><u> </u><u>,</u><u> </u><u>y2</u><u> </u><u>=</u><u> </u><u>-</u><u>7</u>
<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u>








Answer:
14
Step-by-step explanation:
15-14 = 1
1*5=5
5/1=1
D,would be the correct answer.