Write the standard form and index form
of the
the polynomial using variable
from its coefficient Forms [3,15,9,0,-2]
Answers
[tex][/tex]
i. Number of coefficients = 5 ∴ Degree = 5 – 1 = 4 ∴ Index form = 3x4 – 2x3 + 0x2 + 7x + 18 ii. Number of coefficients = 4 ∴ Degree = 4 – 1 = 3 ∴ Index form = 6x3 + x2 + 0x + 7 iii. Number of coefficients = 4 ∴ Degree = 4 – 1 = 3 ∴ Index form = 4x3 + 5x3 – 3x + 3-18-ii-6-1-0
Answer:
i. Number of coefficients = 5 ∴ Degree = 5 – 1 = 4 ∴ Index form = 3x4 – 2x3 + 0x2 + 7x + 18
ii. Number of coefficients = 4 ∴ Degree = 4 – 1 = 3 ∴ Index form = 6x3 + x2 + 0x + 7
iii. Number of coefficients = 4 ∴ Degree = 4 – 1 = 3 ∴ Index form = 4x3 + 5x3 – 3x + 3-18-ii-6-1-0
hope this helps you
Answer:
i. Number of coefficients = 5 ∴ Degree = 5 – 1 = 4 ∴ Index form = 3x4 – 2x3 + 0x2 + 7x + 18
ii. Number of coefficients = 4 ∴ Degree = 4 – 1 = 3 ∴ Index form = 6x3 + x2 + 0x + 7
iii. Number of coefficients = 4 ∴ Degree = 4 – 1 = 3 ∴ Index form = 4x3 + 5x3 – 3x + 3-18-ii-6-1-0
hope this helps you
Answer:
i. Number of coefficients = 5 ∴ Degree = 5 – 1 = 4 ∴ Index form = 3x4 – 2x3 + 0x2 + 7x + 18
ii. Number of coefficients = 4 ∴ Degree = 4 – 1 = 3 ∴ Index form = 6x3 + x2 + 0x + 7
iii. Number of coefficients = 4 ∴ Degree = 4 – 1 = 3 ∴ Index form = 4x3 + 5x3 – 3x + 3-18-ii-6-1-0