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Another homework question - HSN forum # Another homework question

### #1craig1e

• Gender:Male

Posted 09 January 2007 - 08:25 PM

Find constants a,b and c such that (x^2+ax+1)(x^2-3x-2) = x^4-2x^3+bx^2+cx-2

### #2Moondarra

• Location:Fraserburgh
• Interests:Maths, playing guitar and of course music. chess and anything a challenge
• Gender:Male

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

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