Hard Factorisation Problem (1 Viewer)

[frenzal_dude]

How would you factorise this?:
s^3 + 5s^2 + 8s + 4

 

[buchanan]

I wouldn't call it hard.

 

[nonsenseTM]

trial and error using factors of 4, and -1 works thus we can divide the whole thing by x+1

 

[frenzal_dude]

trial and error using factors of 4, and -1 works thus we can divide the whole thing by x+1

how did you know to use factors of 4 and -1, and how'd u know to divide the whole thing by s+1?

I know that since there were no variations in sign, there would be no +ve roots, and when you sub in x = -x, you get 3 variations in sign, so I knew there would be 3 -ve roots (which i now know are -1, -2, -2) but how could you work out those roots without using trial and error?

 

[buchanan]

Just test divisors of the constant term. If one works, say a, then by the Factor Theorem, s-a is a factor. If none work, use a formula.

In your example, -1 and -2 will work, so s+1 and s+2 are factors.

 

[nonsenseTM]

sorry i didn't make it clear, since the constant term is 4 , which comes from the multiplilcation of the roots. so there must a a factor of 4 that is root.

after you find the root using trial and error, which I found was -1, you divide the poly expression by s+1 (factor theorem)

 

[fullonoob]

sorry i didn't make it clear, since the constant term is 4 , which comes from the multiplilcation of the roots. so there must a a factor of 4 that is root.

after you find the root using trial and error, which I found was -1, you divide the poly expression by s+1 (factor theorem)

this is 3u factor theorem, not even really 4u, but used in 4u

 

[shaon0]

How would you factorise this?:
s^3 + 5s^2 + 8s + 4

x^3+5x^2+8x+4
=x^3+x^2+4x^2+8x+4
=x^2(x+1)+4(x^2+2x+1)
=x^2(x+1)+4(x+1)^2
=(x+1)(x^2+4x+4)
=(x+1)(x+2)^2

Try to use the 8 and 4

 

[cutemouse]

Possible rational roots are p/q where p|4 and q|1...

 

[StuartLee]

Hi,

Let P(s) = s^3 + 5s^2 + 8s + 4

P(-1) = (-1)^3 + 5.(-1)^2 + 8.-1 + 4 = -1 + 5 - 8 + 4 = 0

So, s+1 is a factor

So, P(s) = (s + 1)(s^2 +k.s + 4)

coeff of s^2 = 1 + k = 5, so k = 4

P(s) = (s + 1) (s^2 + 4s + 4) = (s+1)(s+2)^2

Stuart