Question

pls.....answer the 29th question...

Answer 1

Let \(f(x)=x^3+3x^2-2\alpha x+\beta\).
Since (x+1) is a factor of f(x), f(-1)=0.
\(f(-1)=(-1)^3+3(-1)^2-2\alpha*(-1)+\beta\)
\(0=-1+3+2\alpha+\beta\)
\(0=2+2\alpha+\beta\)
\(-2=2\alpha+\beta-(1)\)
Since (x+2) is a factor of f(x), f(-2)=0.
\(f(-2)=(-2)^3+3(-2)^2-2\alpha*(-2)+\beta\)
\(0=-8+12+4\alpha+\beta\)
\(0=4+4\alpha+\beta\)
\(-4=4\alpha+\beta-(2)\)
solve the simultaneous equations (1) and (2) to find \(\alpha\) and \(\beta\).
can u do?

Answer 2

no....@ajprincess

Answer 3

sry...need HELP @ajprincess

Answer 4

\(-2=2\alpha+\beta-(1)\)
\(-4=4\alpha+\beta-(2)\)
\((2)-(1)=>-2=2\alpha\)
\(\alpha=-2/2\)
\(\alpha=-1\)
Plug in the value of \(\alpha\) in equation (1) to find \(\beta\)

Answer 5

\[wher did \beta go\i n the calculation\]

Answer 6

@ajprincess

Answer 7

\((2)-(1)=>-4-(-2)=4\alpha+\beta-2\alpha-\beta\)
\(-4+2=2\alpha\)
\(-2=2\alpha\)
\(\alpha=-2/2\)
\(\alpha=-1\)

Answer 8

is it clear nw @basith?

Answer 9

yes..thnx 4 tht.can u help me with the second part of the question also

Answer 10

@ajprincess

Answer 11

jst a minute. am takng a look at it

Answer 12

k

Answer 13

First find the factors of -5.
They are -1,5,1 and -5.
Let f(x)=x^3-3x^2-9x-5
See for which of the values f(x) be zero when u substitute for x.

Answer 14

sry g2g......anyway thnx... :)

Answer 15

that's k. yw:)