Chapter- Factorisation of Polynomials
In each of the following use factor theorem to find whether polynomial g(x) is a factor of
poluomial fx) or not: (1-7)
1. (x) = 6x² + 11x 6; 8(x) = x-3
2 f(x) – 30° +177 +9r? – 77-10; g(x) = x + 5
3. f(x)= x + 3x – x – 3x² + 5x+15, g(x) = x+3
4. f(x) = 6x? – 19.7 + 84, 8(x) = x – 7
5. f(x) = 3r+x? – 20x + 12, 8(x) = 3x – 2
6. f(x) = 2.r? – 9x² + x + 12, g(x) = 3 – 2x
7 f(x) = x – 6×2 +11% -6,8(x) = x2 – 3x + 2
Answer:
1: f(x) = x3 – 6×2 + 11x – 6; g(x) = x – 3
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.
g(x) = x -3 = 0
or x = 3
Remainder = f(3)
Now,
f(3) = (3)3 – 6(3)2 +11 x 3 – 6
= 27 – 54 + 33 – 6
= 60 – 60
= 0
Therefore, g(x) is a factor of f(x)
2: f(x) = 3X4 + 17×3 + 9×2 – 7x – 10; g(x) = x + 5
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.
g(x) = x + 5 = 0, then x = -5
Remainder = f(-5)
Now,
f(3) = 3(-5)4 + 17(-5)3 + 9(-5)2 – 7(-5) – 10
= 3 x 625 + 17 x (-125) + 9 x (25) – 7 x (-5) – 10
= 1875 -2125 + 225 + 35 – 10
= 0
Therefore, g(x) is a factor of f(x).
3: f(x) = x5 + 3×4 – x3 – 3×2 + 5x + 15, g(x) = x + 3
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.
g(x) = x + 3 = 0, then x = -3
Remainder = f(-3)
Now,
f(-3) = (-3)5 + 3(-3)4 – (-3)3 – 3(-3)2 + 5(-3) + 15
= -243 + 3 x 81 -(-27)-3 x 9 + 5(-3) + 15
= -243 +243 + 27-27- 15 + 15
= 0
Therefore, g(x) is a factor of f(x).
4: f(x) = x3 – 6×2 – 19x + 84, g(x) = x – 7
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.
g(x) = x – 7 = 0, then x = 7
Remainder = f(7)
Now,
f(7) = (7)3 – 6(7)2 – 19 x 7 + 84
= 343 – 294 – 133 + 84
= 343 + 84 – 294 – 133
= 0
5: f(x) = 3×3 + x2 – 20x + 12, g(x) = 3x – 2
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.
g(x) = 3x – 2 = 0, then x = 2/3
Remainder = f(2/3)
Now,
f(2/3) = 3(2/3) 3 + (2/3) 2 – 20(2/3) + 12
= 3 x 8/27 + 4/9 – 40/3 + 12
= 8/9 + 4/9 – 40/3 + 12
= 0/9
= 0
6: f(x) = 2×3 – 9×2 + x + 12, g(x) = 3 – 2x
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.
g(x) = 3 – 2x = 0, then x = 3/2
Remainder = f(3/2)
Now,
f(3/2) = 2(3/2)3 – 9(3/2)2 + (3/2) + 12
= 2 x 27/8 – 9 x 9/4 + 3/2 + 12
= 27/4 – 81/4 + 3/2 + 12
= 0/4
= 0
Therefore, g(x) is a factor of f(x).
7: f(x) = x3 – 6×2 + 11x – 6, g(x) = x2 – 3x + 2
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.
g(x) = 0
or x2 – 3x + 2 = 0
x2 – x – 2x + 2 = 0
x(x – 1) – 2(x – 1) = 0
(x – 1) (x – 2) = 0
Therefore x = 1 or x = 2
Now,
f(1) = (1)3 – 6(1)2 + 11(1) – 6 = 1-6+11-6= 12- 12 = 0
f(2) = (2)3 – 6(2)2 + 11(2) – 6 = 8 – 24 + 22 – 6 = 30 – 30 = 0
=> f(1) = 0 and f(2) = 0
Which implies g(x) is factor of f(x).
Question 8: Show that (x – 2), (x + 3) and (x – 4) are factors of x3 – 3×2 – 10x + 24.
Solution:
Let f(x) = x3 – 3×2 – 10x + 24
If x – 2 = 0, then x = 2,
If x + 3 = 0 then x = -3,
and If x – 4 = 0 then x = 4
Now,
f(2) = (2)3 – 3(2)2 – 10 x 2 + 24 = 8 – 12 – 20 + 24 = 32 – 32 = 0
f(-3) = (-3)3 – 3(-3)2 – 10 (-3) + 24 = -27 -27 + 30 + 24 = -54 + 54 = 0
f(4) = (4)3 – 3(4)2 – 10 x 4 + 24 = 64-48 -40 + 24 = 88 – 88 = 0
f(2) = 0
f(-3) = 0
f(4) = 0
Hence (x – 2), (x + 3) and (x – 4) are the factors of f(x)
Question 9: Show that (x + 4), (x – 3) and (x – 7) are factors of x3 – 6×2 – 19x + 84.
Solution:
Let f(x) = x3 – 6×2 – 19x + 84
If x + 4 = 0, then x = -4
If x – 3 = 0, then x = 3
and if x – 7 = 0, then x = 7
Now,
f(-4) = (-4)3 – 6(-4)2 – 19(-4) + 84 = -64 – 96 + 76 + 84 = 160 – 160 = 0
f(-4) = 0
f(3) = (3) 3 – 6(3) 2 – 19 x 3 + 84 = 27 – 54 – 57 + 84 = 111 -111=0
f(3) = 0
f(7) = (7) 3 – 6(7) 2 – 19 x 7 + 84 = 343 – 294 – 133 + 84 = 427 – 427 = 0
f(7) = 0
Hence (x + 4), (x – 3), (x – 7) are the factors of f(x)