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Answers
Question 1: Factorise x3 – 23x2 + 142x – 120
Answer : Let p(x) = x3 – 23x2 + 142x – 120. To begin with, we will start finding the factors of the constant ‘– 120’, which are:
±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60 and ±120
Further, by trial, we find that p(1) = 0. Hence, we conclude that (x – 1) is a factor of p(x). Also, we see that
[x3 – 23x2 + 142x – 120] = x3 – x2 – 22x2 + 22x + 120x – 120
So, by removing the common factors, we have x3 – 23x2 + 142x – 120 = x2(x –1) – 22x(x – 1) + 120(x – 1)
Further, taking ‘x – 1’ common, we get x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
Therefore, x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
Also, note that if we divide p(x) by ‘x – 1’, then the result will be (x2 – 22x + 120)
Going on, x2 – 22x + 120 can be factorised further. So, by splitting the middle term, we get:
x2 – 22x + 120 = x2 – 12x – 10x + 120 … [ (– 12 – 10 = – 22) and {(–12)( –10) = 120}]
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
Therefore, we have x3 – 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12)
Question 2: Factorise x3 – 2x2 – x + 2
Answer : Let p(x) = x3 – 2x2 – x + 2. To begin with, we will start finding the factors of the constant ‘2’, which are: 1, 2
By trial, we find that p(1) = 0. Hence, we conclude that (x – 1) is a factor of p(x).
So, by removing the common factors, we have x3 – 2x2 – x + 2 = x2(x – 2) – (x – 2) = (x2 – 1)(x – 2)
= (x + 1)(x – 1)(x – 2) … [Using the identity (x2 – 1) = (x + 1)(x – 1)]
Therefore, the factors of x3 – 2x2 – x + 2 are (x + 1), (x – 1) and (x – 2)
Example:
Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6.
Solution: To begin with, we know that the zero of the polynomial (x + 2) is –2. Let p(x) = x3 + 3x2 + 5x + 6
Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 = –8 + 12 – 10 + 6 = 0
According to the factor theorem, if p(a) = 0, then (x – a) is a factor of p(x). In this example, p(a) = p(- 2) = 0
Therefore, (x – a) = {x – (-2)} = (x + 2) is a factor of ‘x3 + 3x2 + 5x + 6’ or p(x).
[2]. Example
Factorise 6×2 + 17x + 5 by splitting the middle term.
Solution 1 (By splitting method): As explained above, if we can find two numbers, ‘p’ and ‘q’ such that, p + q = 17 and pq = 6 x 5 = 30, then we can get the factors.
After looking at the factors of 30, we find that numbers ‘2’ and ‘15’ satisfy both the conditions, i.e. p + q = 2 + 15 = 17 and pq = 2 x 15 = 30. So,
6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5
= 6x2 + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)
Therefore, the factors of (6x2 + 17x + 5) are (3x + 1) and (2x + 5) with a remainder, zero.
Question 1: Factorise x3 – 23x2 + 142x – 120
Answer : Let p(x) = x3 – 23x2 + 142x – 120. To begin with, we will start finding the factors of the constant ‘– 120’, which are:
±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60 and ±120
Further, by trial, we find that p(1) = 0. Hence, we conclude that (x – 1) is a factor of p(x). Also, we see that
[x3 – 23x2 + 142x – 120] = x3 – x2 – 22x2 + 22x + 120x – 120
So, by removing the common factors, we have x3 – 23x2 + 142x – 120 = x2(x –1) – 22x(x – 1) + 120(x – 1)
Further, taking ‘x – 1’ common, we get x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
Therefore, x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
Also, note that if we divide p(x) by ‘x – 1’, then the result will be (x2 – 22x + 120)
Going on, x2 – 22x + 120 can be factorised further. So, by splitting the middle term, we get:
x2 – 22x + 120 = x2 – 12x – 10x + 120 … [ (– 12 – 10 = – 22) and {(–12)( –10) = 120}]
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
Therefore, we have x3 – 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12)