Question

Facforcise
P cube -3p Square - 9p - 5​

Answer 1

Step-by-step explanation:

p³-3p²-9p-5

Let the polynomial be f(p)

f(-1)= (-1)³-3(-1)²-9(-1)-5

= -1-3+9-5

= 9-9

= 0

So, (p+1) is a factor of f(p).

p+1 | p³-3p²-9p-5 | p²-4p-5

- p³±p²

-4p²-9p-5

-4p²-4p

-5p-5

-5p-5

0

So, (p+1)(p²-4p-5)= f(p)

= (p+1)(p²-5p+p-5)

= (p+1){p(p-5)+1(p-5)}

= (p+1)(p-5)(p+1)

Answer 2

Answer:

p³- 3p² -9p -5

f(p)=p³-3p²-9p-5

f(-1)=(-1)³-3(-1)²-9(-1)-5

    =-1-3+9-5

    =-9+9=0

therefore -1 is the zero of the polynomial

therefore (x+1) is a factor

by synthetic division

-1   1   -3   -9   -5

     0 -1    4      5

     1   -4    -5    0

therefore 1x²-4x-5                                                                     1 x 5

               1x²-5x+1x-5

               1x(x-5) + 1(x-5)

therefore all the factors are (x+1),(x-5),(1x+1)