the zeroes in 3 0 0 82 0 0 27
those are coefficents of the polynomial
the polynomial is 3x^6+0x^5+0x^4+82x^3+0x^2 + 0x + 27
or, 3 0 0 82 0 0 27
yes, your conclusion is correct. You can prove it by long division.
The book said to replace the x with a zero. Why are two zero's used in the division of this problem. (synthetic division)
21. -3)3 0 0 82 0 0 27
-9 27 -81 -3 9 -27
3 -9 27 1 -3 9 0
Remainder = 0 so that means x+3 is a factor of 3x6+82x3+27
2 answers
From your post I concluded that you had
3x^6 - 82x^3 + 27 ÷ (x+3)
the expression did not contain any x^5 , x^4 , x^2 and x terms so in your synthetic division setup you have to replace these with 0's
that is why your top row looks like
-3 | 3 0 0 82 0 0 27
I assume you know the procedure for synthetic division, the second row is correct
the last row is your "answer row" and since we started with x^6 .... divided by x+3, the answer must start with x^5
so
3x^6 - 82x^3 + 27 ÷ (x+3)
= 3x^5 - 9x^4 + 27x^3 + x^2 - 3x + 9 with zero remainder
and yes, x+3 is a factor since our remainder was zero, the last number in the last row.
3x^6 - 82x^3 + 27 ÷ (x+3)
the expression did not contain any x^5 , x^4 , x^2 and x terms so in your synthetic division setup you have to replace these with 0's
that is why your top row looks like
-3 | 3 0 0 82 0 0 27
I assume you know the procedure for synthetic division, the second row is correct
the last row is your "answer row" and since we started with x^6 .... divided by x+3, the answer must start with x^5
so
3x^6 - 82x^3 + 27 ÷ (x+3)
= 3x^5 - 9x^4 + 27x^3 + x^2 - 3x + 9 with zero remainder
and yes, x+3 is a factor since our remainder was zero, the last number in the last row.
1 answer