<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
4
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
0
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
0
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
8
This implies ;
D ( x ) is not divisible by ( x - 1 )
<h2>Therefore ; </h2>
<h3>B ( x ) & C ( x ) are divisible by ( x - 1 ) </h3>
Answer:
Adding by 4 each time
Step-by-step explanation:
3m - 5p = 12
3m(-3m) -5p = (-3m)+ 12
-5p = -3m + 12
(-5)/-5p = (-5)/ (-3m + 12)
p = 3/5m + 12/-5
Answer:B
Step-by-step explanation: Alternate exterior angles theorem
Option D. D has the matrix of constants [[12], [11], [4]].
Step-by-step explanation:
Step 1:
With the given equations, we can form matrices to represent them.
The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.
Step 2:
The linear system A is represented as
![\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%261%261%5C%5C1%26-11%260%5C%5C1%26-1%264%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\z\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D26%5C%5C17%5C%5C23%5Cend%7Barray%7D%5Cright%5D)
.
Step 3:
The linear system B is represented as
![\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%261%261%5C%5C1%26-11%260%5C%5C1%26-1%2612%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D23%5C%5C17%5C%5C26%5Cend%7Barray%7D%5Cright%5D)
.
Step 4:
The linear system C is represented as
![\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%5C%5C1%26-1%260%5C%5C1%26-1%261%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C11%5C%5C12%5Cend%7Barray%7D%5Cright%5D)
.
Step 5:
The linear system D is represented as
![= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%5C%5C11%5C%5C4%5Cend%7Barray%7D%5Cright%5D)
.
Step 6:
Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.
