Find the values of a and b so that 1 is a root of multiplicty 2 of the polynomial
P(x) = x^873 - 2x^401 - 5x^14 + ax^5 - 3x³ + bx -3
This seems a very messy problem. Is there any easy way to solve this?
Thank you.
P(x) = x^873 - 2x^401 - 5x^14 + ax^5 - 3x³ + bx -3
This seems a very messy problem. Is there any easy way to solve this?
Thank you.
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Yes. For 1 to be a root, P(1) = 0. For it to be a root of multiplicity 2, it must have a max/min at 1, so P'(1)=0.
P(1)=1-2-5+a-3+b-3 = 0, so a+b =12
P'(1) = 873 - 802 - 70 + 5a - 6 + b = 0 so 5a + b = 5
this yields a = -7/4, b = 55/4
I did the work in my head, so double check that, but the principle is sound.
P(1)=1-2-5+a-3+b-3 = 0, so a+b =12
P'(1) = 873 - 802 - 70 + 5a - 6 + b = 0 so 5a + b = 5
this yields a = -7/4, b = 55/4
I did the work in my head, so double check that, but the principle is sound.