function is defined by folliowing rule question.

**thcbender** · December 8th 2009, 09:30 PM

a function is defiend by the following rule:

f (x) is
0 if x lesser than/equal to -3
-1 if -3<x<0
x if x is greater then or equal to 0

find i) F (-3) + f (-2) + f (2)
ii) f (a ^2)

another question:
4 ^ (x+1)/ ( 4 ^(x + 2) – 4 ^ (x) )

and another.

x-1 is a factor of x ^ 3 - (2 X ^ 2) + px + 6
find the value of P

and another
fully factorise the polynomial P(X) = X ^4 - 7x^3 + 12x^2 + 4x -16
given that x = 2 is a double root of P(x)

**acc100jt** · December 8th 2009, 09:58 PM

Originally Posted by thcbender

a function is defiend by the following rule:

f (x) is
0 if x lesser than/equal to -3
-1 if -3<x<0
x if x is greater then or equal to 0

find i) F (-3) + f (-2) + f (2)
ii) f (a ^2)

(i) $f(-3)+f(-2)+f(2)=0+(-1)+2=1$
(ii) Since $a^{2}\geq0$ , so $f(a^{2})=a^{2}$

Originally Posted by thcbender

another question:
4 ^ (x+1)/ ( 4 ^(x + 2) – 4 ^ (x) )

$\frac{4^{x+1}}{4^{x+2}-4^{x}}=\frac{4(4^{x})}{4^{2}4^{x}-4^{x}}$
$=\frac{4(4^{x})}{4^{x}(4^{2}-1)}=\frac{4}{4^{2}-1}=...$

Originally Posted by thcbender

x-1 is a factor of x ^ 3 - (2 X ^ 2) + px + 6
find the value of P

Since $(x-1)$ is a factor, by factor theorem, $f(1)=0$ , where $f(x)=x^{3}-2x^{2}+px+6$
So $1^{3}-2(1)^{2}+p(1)+6=0$ ...solve it you will get your p

Originally Posted by thcbender

fully factorise the polynomial P(X) = X ^4 - 7x^3 + 12x^2 + 4x -16
given that x = 2 is a double root of P(x)

Since $x=2$ is a double root, so $(x-2)^{2}$ is a factor, get the other quadratic factor by either long division, comparison or synthetic division. You will get $f(x)=(x-2)^{2}(x^{2}-3x-4)=(x-2)^{2}(x-4)(x+1)$

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