Polynomials

Exercise 2.4 Part 2

Question 2: Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) `p(x) = 2x^3 + x^2 – 2x – 1`
`g(x) = x + 1`

Answer: Let `g(x) = 0`
Or, `x + 1 = 0`
Or, `x = - 1`

This means -1 is the zero of the given polynomial g(x) = x + 1
Now, according to Factor Theorem, if p(-1) is equal to zero, then g(x) is the factor of `p(x) = 2x^3 + x^2 – 2x – 1`
Or, `p( - 1) = 2( - 1)^3 + ( - 1)^2 – 2( - 1) – 1`
`= 2 x ( - 1) + 1 + 2 – 1`
`= - 2 + 1 + 2 – 1 = 0`
i.e. `p( - 1) = 0`

Since, `p( - 1) = 0`, therefore, `g(x) = (x + 1)` is a factor of `p(x) = 2x^3 + x^2 – 2x - 1`


(ii) `p(x) = x^3 + 3x^2 + 3x + 1`
`g(x) = x + 2`

Answer: Let `g(x) = 0`
Or, `x + 2 = 0`
Or, `x = - 2`, i.e. zero of `(x + 2) = - 2`
Now, `(p – 2) = (- 2)^3 + 3( - 2)^2 + 3( - 2) + 1`
`= - 8 + 12 – 6 + 1 = - 1`

Since, `p(-2) ≠ 0`, therefore, according to Factor Theorem `g(x) = x + 2` is not the factor of given polynomial `p(x) = x^3 + 3x^2 + 3x + 1`


(iii) `p(x) = x^3 – 4x^2 + x + 6`
`g(x) = x – 3`

Answer: Let `g(x) = 0`
Or, `x – 3 = 0`
Or, `x = 3`
Now, `p(3) = 3^3 – 4xx3^2 + 3 + 6`
`= 27 – 36 + 9 = 0`

Since, `p(3) = 0`, therefore, according to Factor Theorem, `g(x) = x + 3` is the factor of given polynomial `p(x) = x^3 – 4x^2 + x + 6`



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