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# Factorize this expression

Wed 13 Jun, 2012 11:02 pm

a^4 - 6a^3 +15a^2 - 18a + 5
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Type: Question • Score: 0 • Views: 746 • Replies: 4
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engineer

1
Thu 14 Jun, 2012 05:41 am
@rishi banerjee,
There are no rational roots to this equation. The only possible rational roots would be -5, -1, 1 and 5 since they must be factors of 5 divided by factors of 1.
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engineer

1
Thu 14 Jun, 2012 07:48 am
@rishi banerjee,
You can factor this into two quadratics and use the quadratic equation to get the roots.

(a^2 - 3a + 1)(a^2 - 3a + 5)
raprap

1
Thu 14 Jun, 2012 08:53 pm
@rishi banerjee,
take a^4 - 6a^3 +15a^2 - 18a + 5
and set equal to (a^2+n1a+m1)(a^2+n2a+m2)

Two things you know right off the bat
m1*m2=5
n1+n2=-6
if this is factorable easily then m1=+/-1 and m2=+/-5
now you also may know that m1+m2+n1n2=15
so set m1=1 and m2=5 so n1n2=9
as n2n2=9 and n1+n2=-6 then n2=-3 and n2=-3

Now you've factored the quartic into two quadratics
a^4 - 6a^3 +15a^2 - 18a + 5 =(a^2-3a+1)(a^2-3a+5)
You can factor this using the quadratic formula

Hint
you will get four real roots
or
two real and two imaginary roots
or
four imaginary roots

Rap

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rishi banerjee

1
Sun 17 Jun, 2012 03:03 am
@engineer,
please can you explain how it came
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