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

### #1

Posted 09 January 2007 - 08:25 PM

### #2

Posted 09 January 2007 - 11:40 PM

(x

ok all you do is multiply out the brackets for the Left Hand Side and then compare

x

once you simplify it down:

x

then just compare both sides:

x

L.H.S = (a - 3)x

R.H.S = - 2x

therefore:

a - 3 = 2

a = 5

now that we've found a we can find b:

L.H.S = -(3a + 1)x

R.H.S = b

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

^{2}+ax+1)(x^{2}-3x-2) = x^{4}-2x^{3}+bx^{2}+cx-2ok all you do is multiply out the brackets for the Left Hand Side and then compare

x

^{4}+ ax^{3}+ x^{2}- 3x^{3}- 3ax^{2}- 3x - 2x^{2}- 2ax - 2once you simplify it down:

x

^{4}+ (a - 3)x^{3}- (3a + 1)x^{2}- (3 + 2a)x - 2then just compare both sides:

x

^{4}+ (a - 3)x^{3}- (3a + 1)x^{2}- (3 + 2a)x - 2 = x^{4}- 2x^{3}+ bx^{2}+ cx - 2L.H.S = (a - 3)x

^{3}R.H.S = - 2x

^{3}therefore:

a - 3 = 2

a = 5

now that we've found a we can find b:

L.H.S = -(3a + 1)x

^{2}R.H.S = b

^{2}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__#### 1 user(s) are reading this topic

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