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Algebra
Factor Theorem
1. Show that ( x 5) and (2 x 1) are factors of 2 x^2 – 11 x + 5.
Solution: p ( x ) = 2 x^2 – 11 x + 5 p (5) = 2 (5)^2 – 11 (5) + 5 = 50 – 55 + 5 = 0
( x – 5) is a factor
p = 2 11 + 5
(2 x – 1) is a factor
2. Show that 2 x + 7 is a factor of 2 x^3 + 7 x^2 4 x – 14.
Solution: Let p ( x ) = 2 x^3 + 7 x^2 – 4 x – 14
p = 2 + 7 4 14
2 x + 7 is a factor
3. Find the value of k , if ( x + 3) is a factor of 3 x^2 + kx + 6.
Solution: Let p ( x ) = 3 x^2 + kx + 6
p (3) = 3 (3)^2 + k (3) + 6 = 0
2
3 2
27 – 3 k + 6 = 0 3 k = 33 k = 11
4. Find the value of a, if ( x a ) is a factor of x^3 – ax^2 + x + 2
Solution: Let p ( x ) = x^3 – ax^2 + x + 2 p ( a ) = a^3 – a^3 + a + 2 = 0 a + 2 = 0 a = 2
5. Find the value of k, if ( x + 2) is a factor of ( x + 1)^7 + 2 x + k****.
Solution: Let p ( x ) = ( x + 1)^7 + 2 x + k
p (2) = (2 + 1)^7 + 2 (2) + k = 0
1 – 4 + k = 0 k = 5
6. Determine the value of a for which the polynomial 2 x^4 – ax^3 + 4 x^2 + 2 x + 1 is divisible by 1 – 2 x****.
Solution: p ( x ) = 2 x^4 – ax^3 + 4 x^2 + 2 x + 1
p = 2 a + 4 + 2
1 – a + 24 = 0
a = 25
1 a 8 8
4 3 2
10. If ( x – 1) and ( x + 1) are factors of x^3 + 3 x^2 + ax + b find a and b****.
Solution: Let p ( x ) = x^3 + 3 x^2 + ax + b
Since ( x 1) is a factor p (1) = 0 (1)^3 + 3(1)^2 + a (1) + b = 0
1 + 3 + a + b = 0 a + b = 4 -----(1) Since ( x + 1) is a factor
p (1) = 0
(1)^3 + 3 (1)^2 + a (1 ) + b = 0
1 + 3 – a + b = 0
a + b = 2 -----(2) Solving (1) and (2)
a + b = 4
a + b = 2
2b = – 6 b = – 3 Sub b = – 3 in equ (2) a + b = 2
a + (–3) = – 2
a = – 2 + 3 a = – 1
b = 3, a = 1
11. If x^2 – 5 x + 6 is a factor of 3 x^3 + ax^2 + b x + 24, find a and b.
Solution: x^2 – 5 x + 6 = 0 ( x – 2) ( x – 3) = 0 Let p ( x ) = 3 x^3 + ax^2 + b x + 24 p (2) = 3 (2)^3 + a (2)^2 + b (2) + 24 = 0
( ( x 2) is a factor)
= 24 + 4 a + 2b + 24 = 0
4 a + 2b = 48
4 a + b = 24 -----(1) p (3) = 3(3)^3 + a (3)^2 + b(3) + 24 = 81 + 9 a + 3b + 24 = 0
[ ( x – 3) is a factor]
9 a + 3b = 105
3 a + b = 35 -----(2)
(1) 2 a + b = 24
(2) 3 a + b = 35
a = 11
a = 11
substitute a = 11 in equation (2)
3 (11) + b = 35
33 + b = 35
b = 35 + 33
b = 2
a = 11 and b = 2
13. If x^3 + ax^2 + b x + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3 find a and b
Solution : Let p ( x ) = x^3 + ax^2 + b x + 6 Since ( x – 2) is a factor p (2) = 0
(2)^3 + a (2)^2 + b(2) + 6 = 0
4 a + 2b = 14
2 a + b = 7 -----(1) Since the remainder is 3 when divided by ( x – 3) p (3) = 3 (3)^3 + a (3)^2 + b (3) + 6 = 3
9 a + 3b = 30
3 a + b = 10 ----(2)
(1) 2 a + b = 7
(2) 3a + b = 10
a = 3 or a = 3 Substitute a = -3 in equation (2)
3a + b = 10
3(3) + b = 10
9 + b = 10
b = 10 + 9 b = 1
a = 3, b = 1
14. Show by factor theorem x – y , y – z and z – x are the factors of
x^2 ( y – z) + y^2 ( z – x ) + z^2 ( x – y ).
Solution:
Let f ( x , y , z ) = x^2 ( y – z ) + y^2 ( z – x ) + z^2 ( x – y ) Put x = y in the above Then RHS = y^2 ( y – z ) + y^2 ( z – y ) + z^2 ( y – y ) = y^3 – y^2 z + y^2 z – y^3 + 0 = 0
( x – y ) is a factor
Similarly ( y – z ), ( y – x ) are also factors.
15. Using factor theorem show that a – b, b – c and c – a are the factors of a (b^2 – c^2 ) + b (c^2 – a^2 ) + c ( a^2 – b^2 ).
Solution: Let f ( a , b , c ) = a (b^2 – c^2 ) + b (c^2 – a^2 ) + c(a^2 – b^2 ) Put a = b in the above = b (b^2 – c^2 ) + b (c^2 – b^2 ) + c (b^2 – b^2 ) = b^3 – bc^2 + bc^2 – b^3 + 0 = 0
( a – b) is a factor. similarly (b – c) (c – a ) are also a factor.
16. If ( x – 2) is a factor of x^2 + ax + b and a + b = 1, find the values of a and b.
Solution: p ( x ) = x^2 + ax + b Since ( x – 2) is a factor p (2) = 0 p (2) = 4 + 2 a + b = 0 2 a + b = 4 -----(1)
a + b = 1 (given) -----(2)
a = 5 from (2) b = 6
