Go To Index
Unit 4: Factorization
Solved Exercise 4.6
1. Find the remainder byusing the remainder theorem when:(i) x3 - 6x2 + 11x - 8 is divided by (x - 1)
Solution:
By the given condition:
⇒ x - 1 = 0
∴ x = 1
Here
⇒ P(x) = x3 - 6x2 + 11x - 8
By substituting value of x, we get:
⇒ P(1) = 13 - 6(12) + 11(1) - 8
⇒ P(1) = 1 - 6 + 11 - 8
⇒ P(1) = 1 + 11 - 6 - 8
⇒ P(1) = 12 - 14
⇒ P(1) = -2
Thus the Remainder is -2 Answer
(ii) x3 + 6x2 + 11x + 8 is divided by (x + 1)
Solution:
By the given condition:
⇒ x + 1 = 0
∴ x = -1
Here
⇒ P(x) = x3 + 6x2 + 11x + 8
By substituting value of x, we get:
⇒ P(-1) = (-1)3 + 6(-12) + 11(-1) + 8
⇒ P(-1) = - 1 + 6 - 11 + 8
⇒ P(-1) = - 1 - 11 + 6 + 8
⇒ P(-1) = - 12 + 14
⇒ P(-1) = 2
Thus the Remainder is 2 Answer
(iii) x3 - x2 - 26 + 40 is divided by (x - 2)
Solution:
By the given condition:
⇒ x - 2 = 0
∴ x = 2
Here
⇒ P(x) = x3 - x2 - 26 + 40
By substituting value of x, we get:
⇒ P(2) = (2)3 - (2)2 - 26 + 40
⇒ P(2) = 8 - 4 + 14
⇒ P(2) = 4 + 14
⇒ P(2) = 18
Thus the Remainder is 18 Answer
(iv) x3 - 3x2 + 4x - 14 is divided by (x + 2)
Solution:
By the given condition:
⇒ x + 2 = 0
∴ x = -2
Here
⇒ P(x) = x3 - 3x2 + 4x - 14
By substituting value of x, we get:
⇒ P(-2) = (-2)3 - 3(-2)2 + 4(-2) - 14
⇒ P(-2) = - 8 - 12 - 8 - 14
⇒ P(-2) = -42
Thus the Remainder is -42 Answer
(v) (2y + 1)3 + 6(3 + 4y) - 9 is divided by (2y + 1)
(vi) 4y3 - 4y + 3 is divided by (2y - 1)
(vii) (2y + 1)3 - 6(3 - 4y) - 10 is divided by (2y - 1)
(viii) x4 + x2y2 + y4 is divided by (x - y)
Solution:
By the given condition:
⇒ x - y = 0
∴ x = y
Here
⇒ P(x) = x4 + x2y2 + y4
By substituting value of x, we get:
⇒ P(y) = (y)4 + (y)2y2 + y4
⇒ P(y) = (y)4 + y2.y2 + y4
⇒ P(y) = y4 + y4 + y4
⇒ P(y) = 3y4
Thus the Remainder is 3y4 Answer
2. Find the value of m, if p(y) = my3 + 4y2 + 3y - 4 and q(y) = y3 - 4y + m leaves the same remainder when divided by (y - 3)
Solution:
By the given condition:
⇒ y - 3 = 0
∴ y = 3
Here
⇒ p(y) = my3 + 4y2 + 3y - 4
& q(y) = y3 - 4y + m
By substituting value of y, we get:
For p(y)
⇒ p(y) = m(3)3 + 4(3)2 + 3(3) - 4
⇒ P(y) = m(27) + 4(9) + 9 - 4
⇒ P(y) = 27m + 36 + 5
⇒ P(y) = 27m + 41 ..... (i)
For q(y)
⇒ q(y) = (3)3 - 4(3) + m
⇒ q(y) = 27 - 12 + m
⇒ q(y) = 15 + m ..... (ii)
Given that both reminders are same, so by equating equation (i) & (ii), we have:
⇒ 27m + 41 = 15 + m
(by subtracted m and 41 on both side)
⇒ 27m + 41 - m - 41 = 15 + m - m - 41
⇒ 27m - m +
⇒ 26m = - 26
⇒ m = -
⇒ m = -1
Answer: Thus the value of m is -1
3. If the polynomial 4x3 - 7x2 + 6x - 3k is exactly divisible by (y + 2), find the value of k
Solution:
By the given condition:
⇒ x + 2 = 0
∴ x = -2
Here
⇒ p(x) = 4x3 - 7x2 + 6x - 3k
By substituting value of x, we get:
⇒ p(-2) = 4(-2)3 - 7(-2)2 + 6(-2) - 3k
⇒ p(-2) = 4(-8) - 7(4) - 12 - 3k
⇒ p(-2) = - 32 - 28 - 12 - 3k
⇒ p(-2) = - 72 - 3k ..... (i)
For exactly division
⇒ p(-2) = 0 .....(ii)
By equating equation (i) & (ii), we get:
⇒ - 72 - 3k = 0
⇒ - 72 + 72 - 3k = 0 + 72
⇒ -
⇒ -3k = 72 (Divide both side by -3)
⇒
⇒ k = -24
Answer: Thus the value of k is -24
4. Find the value of r, if (y + 2) is a factor of the polynomial 3y2 - 4ry - 4r2.
Solution:
By the given condition:
⇒ y + 2 = 0
∴ x = -2
Here
⇒ p(y) = 3y2 - 4ry - 4r2
& p(-2) = 0
Then 0 = 3y2 - 4ry - 4r2
By substituting value of y, we get:
⇒ 0 = 3(-2)2 - 4r(-2) - 4r2
⇒ 0 = 3(4) + 8r - 4r2
⇒ 0 = 12 + 8r - 4r2
⇒ 0 = -4(-3 - 2r + r2)
(Divide both side by -4)
⇒ 0 / -4 =
OR r2 - 2r - 3 = 0 (by arranging in descending order)
(It can be factorize by breaking method)
Rough Work:
As last term has minus sign so middle term is obtained by subtracting two numbers which are factors of last term i.e. 3
Factor of 3 = 1, 3
∴ 3 - 1 = 2
As middle term has minus sign so greater number contain minus sign
⇒ r(r - 3) + 1(r - 3) = 0
⇒ (r - 3)(r + 1) = 0
Let
⇒ r - 3 = 0 ⇒ r = 0 + 3 ⇒ r = 3 | ⇒ r + 1 = 0 ⇒ r = 0 - 1 ⇒ r = -1 |
Answer: Thus the value of r is 3 and -1.
No comments:
Post a Comment