factor theorem.

• Jun 25th 2006, 08:46 AM
kwtolley
factor theorem.
Let f(x)=x^3-8x^2+17x-9. Use the factor theorem to find other solutions to f(x)-f(1)=0, besides x=1. My answer is 2,5 could that be right. thanks for looking.
• Jun 25th 2006, 08:13 PM
malaygoel
Quote:

Originally Posted by kwtolley
Let f(x)=x^3-8x^2+17x-9. Use the factor theorem to find other solutions to f(x)-f(1)=0, besides x=1. My answer is 2,5 could that be right. thanks for looking.

$f(x)=x^3-8x^2+17x-9$
$f(x)=f(1)=1-8+17-9=1$
$f(x)=x^3-8x^2+17x-9=1$
$f(x)=x^3-8x^2+17x-10=0$
Yes, 2 and 5 are solutions of this equation besides 1.

KeepSmiling
Malay
• Jun 25th 2006, 08:49 PM
Soroban
Hello, kwtolley!

Another approach . . .

Quote:

Let $f(x) \:=\:x^3 - 8x^2 + 17x - 9$
Use the factor theorem to find other solutions to $f(x)-f(1)\,=\,0$, besides $x=1.$
My answers are: $2,\;5.$
We are given: . $f(x) - f(1) \:= \:0$

]. . . . . $(x^3 - 8x^2 + 17x - 9) - (1^3 - 8\cdot1^2 + 17\cdot1 - 9) \;= \;0$

. . . . . . . . . $(x^3 - 1^3) - 8(x^2 - 1^2) + 17(x - 1) - 9 + 9 \;= \;0$

. . . $(x - 1)(x^2 + x + 1) - 8(x - 1)(x + 1) + 17(x - 1) \;= \;0$

. . . . . . . . . . . . . . . $(x - 1)(x^2 + x + 1 - 8[x+1] + 17) \;= \;0$

. . . . . . . . . . . . . . . . . . . . . . . . . $(x - 1)(x^2 - 7x + 10) \;= \;0$

. . . . . . . . . . . . . . . . . . . . . . . . . $(x - 1)(x - 2)(x - 5) \;= \;0$

Therefore, the solutions are: . $x \:= \:1,\;2,\;5$
• Jun 25th 2006, 10:10 PM
malaygoel
Quote:

Originally Posted by Soroban
Hello, kwtolley!

Another approach . . .

We are given: . $f(x) - f(1) \:= \:0$

]. . . . . $(x^3 - 8x^2 + 17x - 9) - (1^3 - 8\cdot1^2 + 17\cdot1 - 9) \;= \;0$

. . . . . . . . . $(x^3 - 1^3) - 8(x^2 - 1^2) + 17(x - 1) - 9 + 9 \;= \;0$

. . . $(x - 1)(x^2 + x + 1) - 8(x - 1)(x + 1) + 17(x - 1) \;= \;0$

. . . . . . . . . . . . . . . $(x - 1)(x^2 + x + 1 - 8[x+1] + 17) \;= \;0$

. . . . . . . . . . . . . . . . . . . . . . . . . $(x - 1)(x^2 - 7x + 10) \;= \;0$

. . . . . . . . . . . . . . . . . . . . . . . . . $(x - 1)(x - 2)(x - 5) \;= \;0$

Therefore, the solutions are: . $x \:= \:1,\;2,\;5$

Great.
Alternative approach are always interesting.

KeepSmiling
Malay
• Jun 26th 2006, 09:18 PM
kwtolley
Factor theorem
Thanks to everyone for looking it over with me.