ATAR Notes: Forum
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
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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
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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
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oh! i never thought abought synthetic i tried sub in P(-1/2)
thank you
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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
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LOL i feel so stupid when its explained, thank you!