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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: 1i1ii1i on February 15, 2012, 10:23:39 pm

Title: Remainder theorem Question
Post by: 1i1ii1i on February 15, 2012, 10:23:39 pm

When the polynomial P(x) =kx^3-6x^2+kx+3 is divided by 2x+1 the remainder is -1, what is the value of k

Title: Re: Remainder theorem Question
Post by: Special At Specialist on February 15, 2012, 10:45:45 pm

Use long/synthetic division to divide the polynomial by 2x + 1:

-1/2  [k     -6         k                 3]
        [       -k/2      3 + k/4      -5k/8 - 3/2]
         k   -6 - k/2    5k/4 + 3    -5k/8 + 3/2

Since the remainder is -1, then:
-5k/8 + 3/2 = -1
5k/8 = 5/2
k/8 = 1/2
k = 4

Title: Re: Remainder theorem Question
Post by: 1i1ii1i on February 15, 2012, 10:49:15 pm

oh! i never thought abought synthetic i tried sub in P(-1/2)
thank you

Title: Re: Remainder theorem Question
Post by: Panicmode on February 15, 2012, 10:59:59 pm

Orrrr... you could use the remainder theorem.

P(-1/2) = -1

-1 = k(-1/2)^3 - 6(-1/2)^2 + k(-1/2) + 3
-4 = -k/8 - 3/2 - k/2
-5/2 = - (k + 4k)/8
20 = 5k
k = 4

Title: Re: Remainder theorem Question
Post by: 1i1ii1i on February 16, 2012, 08:41:15 pm

LOL i feel so stupid when its explained, thank you!