I believe 5 different ways. The options being two packs of 9-one pack of 9 and 3 packs of 3- 1 pack of 9, 2 packs of 3 and 3 singles- one pack of 9, one pack of 3, and 6 singles- and then one pack of 9 and 9 singles.
Hope this helped!

8 0

2 years ago

PLZ HELP QUICK-------------->

sveticcg [70]

<h2>>>> Answer <<<</h2>

Let's check which polynomial is divisible by ( x - 1 ) using hit , trial and error method .

A ( x ) = 3x³ + 2x² - x

The word " divisible " itself says that " it is a factor "

Using factor theorem ;

Let;

=> x - 1 = 0

=> x = 1

Substitute the value of x in p ( x )

p ( 1 ) =

3 ( 1 )³ + 2 ( 1 )² - 1

3 ( 1 ) + 2 ( 1 ) - 1

3 + 2 - 1

5 - 1

This implies ;

A ( x ) is not divisible by ( x - 1 )

Similarly,

B ( x ) = 5x³ - 4x² - x

B ( 1 ) =

5 ( 1 )³ - 4 ( 1 )² - 1

5 ( 1 ) - 4 ( 1 ) - 1

5 - 4 - 1

5 - 5

This implies ;

B ( x ) is divisible by ( x - 1 )

Similarly,

C ( x ) = 2x³ - 3x² + 2x - 1

C ( 1 ) =

2 ( 1 )³ - 3 ( 1 )² + 2 ( 1 ) - 1

2 ( 1 ) - 3 ( 1 ) + 2 - 1

2 - 3 + 2 - 1

4 - 4

This implies ;

C ( x ) is divisible by ( x - 1 )

Similarly,

D ( x ) = x³ + 2x² + 3x + 2

D ( 1 ) =

( 1 )³ + 2 ( 1 )² + 3 ( 1 ) + 2

1 + 2 + 3 + 2

This implies ;

D ( x ) is not divisible by ( x - 1 )

<h2>Therefore ; </h2>

<h3>B ( x ) & C ( x ) are divisible by ( x - 1 ) </h3>

6 0

3 years ago