Please answer this question.
Attachments:
Answers
Answered by
1
Answer :-
1.
f(x) = x³-ax²+6x-a
f(x) divided by x-a,
x-a = 0
x = a
f(a) = a³-a³+6a-a = 5a
5a is the reminder when f(x) is divided by x-a
••••••••••
2.
f(x) = x³+10x²+ax+b
f(1) = 1+10+a+b = 0
a+b+11 = 0 (1)
f(-2) = -8+40-2a+b = 0
2a-b-32 = 0 (2)
By elimination, (2) + (1)
3a = 21
a = 7 & b = -18
••••••••••
3. f(x) = x³+3x²+4x+a
x+6 is a factor of f(x), x = -6
f(-6) = (-6)³+3(-6)²+4(-6)+a = 0
-216+108-24+a = 0
a = 240-108
a = 132
••••••••••
4. Let f(x) = xn+a n
In order to prove that x+a is a factor of f(x) for any odd positive integer n, it is sufficient to show that f(−a)=0.
f(−a) = (−a)n+an = (−1)n an+an
f(−a)=(−1+1)an [n is odd positive integer]
f(−a) = 0 × an = 0
Hence, x+a is a factor of xn+an,
when n is an odd positive integer.
5.f(x) = x^6-px^5+x^4-px³+3x-p+2
f(p)=p^6-p^6+p^4-p^4+3p-p+2=0
2p = -2
p = -1
••••••••••
6. f(x) = x³-8x²+8x+21
if 3 is zero of f(x), f(3) = 0
f(3) = (3)³-8(3)²+8(3)+21
= 27 - 72 + 24 + 21
= 72 - 72
= 0
••••••••••
7. f(x) = x²+2x+3
In f(x), a = 1 ; b = 2 ; c = 3
Discriminant of f(x),
x = b² - 4ac
= 4 - 4(3)
= -8 < 0
Discriminant of f(x) is less than zero.
Then f(x) has no zeroes.
••••••••••
8. f(x) = 2x³+5x²-5x-1
Integral zeroes of f(x), prime factor of constant term is ±1.
f(1) = 2(1)³+5(1)²-5(1)-1
= 2+5-5-1
= 1 ≠ 0
f(-1) = 2(-1)³+5(-1)²-5(-1)-1
= -2-5+5-1
= -3 ≠ 0
f(x) has no integral zeroes.
••••••••••
9. f(x) = x³-ax²-13x+b
x-1 is a factor of f(x),
f(1) = 1-a-13+b = 0
a-b+12 = 0 (1)
x+3 is a factor of f(x),
f(-3) = (-3)³-a(-3)²-13(-3)+b = 0
-27-9a+39+b = 0
9a-b-12 = 0 (2)
By elimination, (2) - (1)
8a = 24
a = 3 & b = 15
••••••••••
10. f(x) = x³+mx²+nx+6
x-2 is a factor of f(x), f(2) = 0
(2)³+m(2)²+n(2)+6 = 0
8+4m+2n+6 = 0
2m+n+7 = 0 (1)
when f(x) is divided by x-3, reminder is 3.
f(3) + 3 = 0
(3)³+m(3)²+n(3)+6+3 = 0
27+9m+3n+9 = 0
9m+3n+36 = 0
3m+n+13 = 0 (2)
By elimination, (2) - (1)
m = 6 & n = -19
••••••••••
11. f(x) = x^4+px³+2x²-3x+q
x±1 is a factor of f(x), f (±1) = 0
f (1) = 1+p+2-3+q = 0
p+q = 0 (1)
f(-1) = 1-p+2+3+q = 0
p-q = 6 (2)
By elimination,
p = 3 & q = -3
••••••••••
12. f(x) = ax^4+2x³-3x²+bx-4
x+2 and I x-2 are the factors of f(x),
f(2) = 0
a(2)^4+2(2)³-3(2)²+b(2)-4 = 0
16a+16-12+2b-4 = 0
8a+b = 0 (1)
f(-2) = 0
16a-16-12-2b-4 = 0
16a-2b-32 = 0
8a-b = 16 (2)
By elimination, (2) + (1)
a = 1 & b = -8
✷••••••••••✷
Similar questions