Maths....Polynomial with ans

Q4)
Given a polynomail: P(x)=ax^3-bx^2+cx-8
and 3 conditions:
2 is a zero , that means P(2) =0 ...(1)
-1 is a zero , that means P(-1)=0 ...(2)
P(3)=28 ...(3)

from (1) we have
a(2)^3-b(2)^2+c(2)-8=0
8a-4b+2c-8=0
4a-2b+c=4 ... (4)

from (2) we have
a(-1)^3-b(-1)^2+c(-1)-8=0
-a-b-c-8=0
a+b+c=-8 ... (5)

from (3) we have
a(3)^3-b(3)^2+c(3)-8=28
27a-9b+3c-8=28
27a-9b+3c=36
9a-3b+c=12...(6)

from (5) we know c=-a-b-8...(7)

sub (7) to (4) and (6) respectively
(4) becomes:
4a-2b+(-a-b-8)=4
4a-2b-a-b-8=4
3a-3b=12
a-b=4 ...(8)

(6) becomes:
9a-3b+(-a-b-8)=12
9a-3b-a-b-8=12
8a-4b=20
2a-b=5... (9)

solving (8) and (9) simultaneously
(9)-(8): a=1 ...(10)

sub (10) into (8)
1-b=4, so b=-3 ...(11)

sub (10) and (11) into (7),
c=-(1)-(-3)-8
=-6

Q5) (b)
because we know 4 is a root
and actually, 4 is a repeated root
hence (x-4) and (x-4) are two factors
of the polynomial

by looking at the constant term of the
polynomial: 16 we know the remaining
factor is something like (x+1)

then we calculate P(-1), it happens to be 0
hence, by factor theorm, (x+1) is another factor
so P(x) = (x+1)(X-4)^2

10) (b)
let P(x) = x^5-9x^4+ux^3-18x^2+v+27
differentiate P(x) w.r.t. x
P'(x)= 5x^4 -36x^3+3ux^2-36x

we have two given condtion:
P(3)=0 ...(1)
P'(3)=0 ...(2)

consider (2) first
5(3)^4 -36(3)^3 +3u(3)^2 -36(3)=0
405 - 324 +27u -108=0
u=1 ...(3)

consider (1) and sub. (3) into it
(3)^5-9(3)^4+(1)(3)^3-18(3)^2+v+27=0
324 -729 +27 -162 +v +27=0
v=513

4) zeros mean roots
P(2) = 8a - 4b ＋ 2c - 8 = 0
P(-1) = -a - b - c - 8 = 0
P(3) = 27a - 9b ＋ 3c = 28
then sovle the 3 equ 3 unknown by any method you like such as elimination

5(b) no need step, just write down the answer
you may try to do division to get the other factor i.e.
　　　　　　　x ＋ 1
　　　　　　 ---------------------------------
(x^2 - 8x＋16) / x^3 - 7x^2 ＋ 8x ＋ 16
　　　　　-)　 x^3 - 8x^2 ＋ 16x
　　　　　　________________________
　　　　　　　　　 x^2　-　8x ＋ 16
　　　　　　　　　 x^2　-　8x ＋ 16
　　　　　　________________________


10b) let P(x) = x^5 - 9x^4 ＋ux^3 - 18x^2 ＋ v＋27
given P(x) = 0 has a triple root at x =3,
=> P(3) = 0 and P'(3) = 0
P'(3)=0
=> 405 ＋ 972＋ 27u -108 =0
=> u=-47
P(3) = 0
=>243 - 729 ＋ 27u - 162 ＋ v 27 =0
=> v = 621-27u = 1890