Factorise:
(i) x3 − 2x2 − x + 2 (ii) x3 + 3x2 −9x − 5
(iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 − 2y − 1
Factorise:
(i) x3 − 2x2 − x + 2 (ii) x3 + 3x2 −9x − 5
(iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 − 2y − 1
(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)
(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)
