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Determine which of the following polynomials has (x + 1) a factor : (i) (ii)
(iii)
(iv)
(i) If (x + 1) is a factor of p(x) = x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).
p(x) = x3 + x2 + x + 1
p(−1) = (−1)3 + (−1)2 + (−1) + 1
= − 1 + 1 − 1 + 1 = 0
Hence, x + 1 is a factor of this polynomial.
(ii) If (x + 1) is a factor of p(x) = x4 + x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).
p(x) = x4 + x3 + x2 + x + 1
p(−1) = (−1)4 + (−1)3 + (−1)2 + (−1) + 1
= 1 − 1 + 1 −1 + 1 = 1
As p(− 1) ≠ 0,
Therefore, x + 1 is not a factor of this polynomial.
(iii) If (x + 1) is a factor of polynomial p(x) = x4 + 3x3 + 3x2 + x + 1, then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.
p(−1) = (−1)4 + 3(−1)3 + 3(−1)2 + (−1) + 1
= 1 − 3 + 3 − 1 + 1 = 1
As p(−1) ≠ 0,
Therefore, x + 1 is not a factor of this polynomial.
(iv) If(x + 1) is a factor of polynomial p(x) = 
, then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.
As p(−1) ≠ 0,
Therefore, (x + 1) is not a factor of this polynomial.
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the
following cases: (i) p(x) = (ii)
(iii) p(x) =
(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then p(−1) must be zero.
p(x) = 2x3 + x2 − 2x − 1
p(−1) = 2(−1)3 + (−1)2 − 2(−1) − 1
= 2(−1) + 1 + 2 − 1 = 0
Hence, g(x) = x + 1 is a factor of the given polynomial.
(ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then p(−2) must
be 0.
p(x) = x3 +3x2 + 3x + 1
p(−2) = (−2)3 + 3(−2)2 + 3(−2) + 1
= − 8 + 12 − 6 + 1
= −1
As p(−2) ≠ 0,
Hence, g(x) = x + 2 is not a factor of the given polynomial.
(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then p(3) must
be 0.
p(x) = x3 − 4 x2 + x + 6
p(3) = (3)3 − 4(3)2 + 3 + 6
= 27 − 36 + 9 = 0
Hence, g(x) = x − 3 is a factor of the given polynomial.
Find the value of k, if x – 1 is a factor of p(x) in each of the following cases: (i) p(x) = (ii) p(x) =
(iii) p(x) =
(iv) p(x) =
If x − 1 is a factor of polynomial p(x), then p(1) must be 0.
(i) p(x) = x2 + x + k
p(1) = 0
⇒ (1)2 + 1 + k = 0
⇒ 2 + k = 0
⇒ k = −2
Therefore, the value of k is −2.
(ii) 
p(1) = 0
(iii) 
p(1) = 0
(iv) p(x) = kx2 − 3x + k
⇒ p(1) = 0
⇒ k(1)2 − 3(1) + k = 0
⇒ k − 3 + k = 0
⇒ 2k − 3 = 0
Therefore, the value of k is
.
Factorise : (i) (ii)
(iii)
(iv)
(i) 12x2 − 7x + 1
We can find two numbers such that pq = 12 × 1 = 12 and p + q = −7. They are p = −4 and q = −3.
Here, 12x2 − 7x + 1 = 12x2 − 4x − 3x + 1
= 4x (3x − 1) − 1 (3x − 1)
= (3x − 1) (4x − 1)
(ii) 2x2 + 7x + 3
We can find two numbers such that pq = 2 × 3 = 6 and p + q = 7.
They are p = 6 and q = 1.
Here, 2x2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
= (x + 3) (2x+ 1)
(iii) 6x2 + 5x − 6
We can find two numbers such that pq = −36 and p + q = 5.
They are p = 9 and q = −4.
Here,
6x2 + 5x − 6 = 6x2 + 9x − 4x − 6
= 3x (2x + 3) − 2 (2x + 3)
= (2x + 3) (3x − 2)
(iv) 3x2 − x − 4
We can find two numbers such that pq = 3 × (− 4) = −12
and p + q = −1.
They are p = −4 and q = 3.
Here,
3x2 − x − 4 = 3x2 − 4x + 3x − 4
= x (3x − 4) + 1 (3x − 4)
= (3x − 4) (x + 1)
Factorise : (i) (ii)
(iii)
(iv)
(i) Let p(x) = x3 − 2x2 − x + 2
All the factors of 2 have to be considered. These are ± 1, ± 2.
By trial method,
p(−1) = (−1)3 − 2(−1)2 − (−1) + 2
= −1 − 2 + 1 + 2 = 0
Therefore, (x +1 ) is factor of polynomial p(x).
Let us find the quotient on dividing x3 − 2x2 − x + 2 by x + 1.
By long division,
It is known that,
Dividend = Divisor × Quotient + Remainder
∴ x3 − 2x2 − x + 2 = (x + 1) (x2 − 3x + 2) + 0
= (x + 1) [x2 − 2x − x + 2]
= (x + 1) [x (x − 2) − 1 (x − 2)]
= (x + 1) (x − 1) (x − 2)
= (x − 2) (x − 1) (x + 1)
(ii) Let p(x) = x3 − 3x2 − 9x − 5
All the factors of 5 have to be considered. These are ±1, ± 5.
By trial method,
p(−1) = (−1)3 − 3(−1)2 − 9(−1) − 5
= − 1 − 3 + 9 − 5 = 0
Therefore, x + 1 is a factor of this polynomial.
Let us find the quotient on dividing x3 + 3x2 − 9x − 5 by x + 1.
By long division,
It is known that,
Dividend = Divisor × Quotient + Remainder
∴ x3 − 3x2 − 9x − 5 = (x + 1) (x2 − 4x − 5) + 0
= (x + 1) (x2 − 5x + x − 5)
= (x + 1) [(x (x − 5) +1 (x − 5)]
= (x + 1) (x − 5) (x + 1)
= (x − 5) (x + 1) (x + 1)
(iii) Let p(x) = x3 + 13x2 + 32x + 20
All the factors of 20 have to be considered. Some of them are ±1,
± 2, ± 4, ± 5 ……
By trial method,
p(−1) = (−1)3 + 13(−1)2 + 32(−1) + 20
= − 1 +13 − 32 + 20
= 33 − 33 = 0
As p(−1) is zero, therefore, x + 1 is a factor of this polynomial p(x).
Let us find the quotient on dividing x3 + 13x2 + 32x + 20 by (x + 1).
By long division,
It is known that,
Dividend = Divisor × Quotient + Remainder
x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0
= (x + 1) (x2 + 10x + 2x + 20)
= (x + 1) [x (x + 10) + 2 (x + 10)]
= (x + 1) (x + 10) (x + 2)
= (x + 1) (x + 2) (x + 10)
(iv) Let p(y) = 2y3 + y2 − 2y − 1
By trial method,
p(1) = 2 ( 1)3 + (1)2 − 2( 1) − 1
= 2 + 1 − 2 − 1= 0
Therefore, y − 1 is a factor of this polynomial.
Let us find the quotient on dividing 2y3 + y2 − 2y − 1 by y − 1.
p(y) = 2y3 + y2 − 2y − 1
= (y − 1) (2y2 +3y + 1)
= (y − 1) (2y2 +2y + y +1)
= (y − 1) [2y (y + 1) + 1 (y + 1)]
= (y − 1) (y + 1) (2y + 1)
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