344 thousands× 10 1 =
1 answer:
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Answer:
None of these.
Step-by-step explanation:
Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.
Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.
In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.
If we look at the first term x^3 and we divide it by 13 we get we cannot reduce it so it is not a fraction so 13 is not a factor of P
None of these is the right option.