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Remainder Theorm, i'm stuck =(

I really can't work out this question, could anyone help me?

f(x) = x^3 + ax^2 + bx + 6

When f(x) is divided by (x-1) the remainder is 24,
When f(x) is divided by (x+2) the remainder is also 24.

Use the Remadiner theorm to fine the numberical values of a and b.


I just don't understand how to work it out =( Thanks everyone.
Reply 1
Avatar for steve2005
steve2005
17
RichardF
I really can't work out this question, could anyone help me?

f(x) = x^3 + ax^2 + bx + 6

When f(x) is divided by (x-1) the remainder is 24,
When f(x) is divided by (x+2) the remainder is also 24.

Use the Remadiner theorm to fine the numberical values of a and b.


I just don't understand how to work it out =( Thanks everyone.

f(1)=24f(1)=24
f(−2)=24f(-2)=24
(1)3+a(1)2+b(1)+6=24⇒a+b=17(1)^3 + a(1)^2 + b(1) + 6 = 24 \Rightarrow a+b = 17

(−2)3+a(−2)2+b(−2)+6=24⇒4a−2b=26(-2)^3 + a(-2)^2 + b(-2) + 6 = 24 \Rightarrow 4a-2b = 26

...... Just finish off to get aa and bb.
Reply 3
Avatar for RichardF
RichardF
OP
Thanks so much all!! i got it now =)
Reply 4
Avatar for JamesyB
JamesyB
12
Quadrifolium
(1)3+a(1)2+b(1)+6=24⇒a+b=17(1)^3 + a(1)^2 + b(1) + 6 = 24 \Rightarrow a+b = 17

(−2)3+a(−2)2+b(−2)+6=24⇒4a−2b=26(-2)^3 + a(-2)^2 + b(-2) + 6 = 24 \Rightarrow 4a-2b = 26

...... Just finish off to get aa and bb.


Exactly this.

Out of Interest what does your sig represent??
JamesyB
Out of Interest what does your sig represent??
138+∑n=0∞(−1)(n+1)(2n+1)!(n+2)!n!4(2n+3)=1+52=ϕ.[br]\displaystyle\frac{13}{8}+\sum_{n=0}^{\infty}\frac{\left(-1\right)^{\left(n+1\right)}(2n+1)!}{(n+2)!n!4^{(2n+3)}} = \dfrac{1+\sqrt{5}}{2} = \phi. [br]

It's an infinite series for ϕ\phi. See here and here.
Reply 6
Avatar for JamesyB
JamesyB
12
Quadrifolium
138+∑n=0∞(−1)(n+1)(2n+1)!(n+2)!n!4(2n+3)=1+52=ϕ.[br]\displaystyle\frac{13}{8}+\sum_{n=0}^{\infty}\frac{\left(-1\right)^{\left(n+1\right)}(2n+1)!}{(n+2)!n!4^{(2n+3)}} = \dfrac{1+\sqrt{5}}{2} = \phi. [br]

It's an infinite series for ϕ\phi. See here and here.


Very Nice :smile: especially the infinity + 1 :biggrin:

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