Find constants a,b and c such that (x^2+ax+1)(x^2-3x-2) = x^4-2x^3+bx^2+cx-2
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Another homework question
Started by craig1e, Jan 09 2007 08:25 PM
1 reply to this topic
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Posted 09 January 2007 - 11:40 PM
(x2+ax+1)(x2-3x-2) = x4-2x3+bx2+cx-2
ok all you do is multiply out the brackets for the Left Hand Side and then compare
x4 + ax3 + x2 - 3x3 - 3ax2 - 3x - 2x2 - 2ax - 2
once you simplify it down:
x4 + (a - 3)x3 - (3a + 1)x2 - (3 + 2a)x - 2
then just compare both sides:
x4 + (a - 3)x3 - (3a + 1)x2 - (3 + 2a)x - 2 = x4 - 2x3 + bx2 + cx - 2
L.H.S = (a - 3)x3
R.H.S = - 2x3
therefore:
a - 3 = 2
a = 5
now that we've found a we can find b:
L.H.S = -(3a + 1)x2
R.H.S = b2
therefore:
-(3a + 1) = b
-(15 + 1) = b
b = -16
now do the same for c:
L.H.S = -(3 + 2a)x
R.H.S = cx
therefore:
-(3 + 2a) = c
-(3 + 10) = c
c = -13
ANS: a = 5, b = -16, c = -13
ok all you do is multiply out the brackets for the Left Hand Side and then compare
x4 + ax3 + x2 - 3x3 - 3ax2 - 3x - 2x2 - 2ax - 2
once you simplify it down:
x4 + (a - 3)x3 - (3a + 1)x2 - (3 + 2a)x - 2
then just compare both sides:
x4 + (a - 3)x3 - (3a + 1)x2 - (3 + 2a)x - 2 = x4 - 2x3 + bx2 + cx - 2
L.H.S = (a - 3)x3
R.H.S = - 2x3
therefore:
a - 3 = 2
a = 5
now that we've found a we can find b:
L.H.S = -(3a + 1)x2
R.H.S = b2
therefore:
-(3a + 1) = b
-(15 + 1) = b
b = -16
now do the same for c:
L.H.S = -(3 + 2a)x
R.H.S = cx
therefore:
-(3 + 2a) = c
-(3 + 10) = c
c = -13
ANS: a = 5, b = -16, c = -13
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