Question

3z²-13z+4 is divided by (z-2) then the remainder is​

Answer 1

Step-by-step explanation:

\begin{gathered}z - 3 = 0 \\ z = 3 \\az + 4 {z}^{3} + 4 {z}^{2} + 3z - 4 \\ a \times 3 + 4( {3}^{3} ) + 4 {}^{2} + 3 \times 3- 4 \\ 3a + 108 + 16 + 9 - 4 \\ 3a + 129.....1\end{gathered}

z−3=0

z=3

az+4z

3

+4z

2

+3z−4

a×3+4(3

3

)+4

2

+3×3−4

3a+108+16+9−4

3a+129.....1

\begin{gathered}z = 3 \\ {z}^{3} - 4z + a \\ (3) {}^{3} - 4 \times 3 + a \\ 27 - 12 + a \\ 15 + a....2\end{gathered}

z=3

z

3

−4z+a

(3)

3

−4×3+a

27−12+a

15+a....2

Since, 1=2

\begin{gathered}3a + 129 = 15 + a \\ 2a = - 114 \\ a = - 57\end{gathered}

3a+129=15+a

2a=−114

a=−57

Hope this helps you :-)))))

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Answer 2

Answer:

10 is the reminder

Step-by-step explanation:

see the answer in picture