Question 1. Write the numerical co-efficient and degree of each term of:
Solution:
Term | Numerical co-efficient | Degree |
x/2 | 1/2 | 1 |
-3x^{2} | -3 | 2 |
5/2 | 3 | |
-5x^{4} | -5 | 4 |
Question 2. Find the remainder when x^{3} â€“ ax^{2} + 4x â€“ a is divided by (x â€“ a).
Solution: p(x) = x^{3 }â€“ ax^{2} + 4x â€“ a (x â€“ a) = 0
â‡’ x = a
âˆ´ p(a) = (a)^{3} â€“ a(a)^{2} + 4(a) â€“ a = a^{3} â€“ a^{3 }+ 4a â€“ a = 4a â€“ a = 3a
âˆ´ The required remainder = 3a
Question 3. When the polynomial kx^{3} + 9x^{2} + 4x â€“ 8 is divided x + 3, then a remainder 7 is obtained.
Find the value of k.
Solution: Here, p(x) = kx^{3} + 9x^{2 }+ 4x â€“ 8
Since, Divisor = x + 3
âˆ´ x + 3 = 0
â‡’ x = â€“3
âˆ´ p(â€“3) = 7
â‡’ k(â€“3)^{3} + 9(â€“3)^{2 }+ 4(â€“3) â€“ 8 = 7
â‡’ â€“27k + 81 â€“ 12 â€“ 8 = 7
â‡’ â€“27k = 7 â€“ 81 + 12 + 8
â‡’ â€“27k = 27 â€“ 81
â‡’ â€“27k = â€“54
â‡’ k= -(54/27) = 2
Thus, the required value of k = 2.
Question 4. (a) For what value of k, the polynomial x^{2} + (4 â€“ k)x + 2 is divisible by x â€“ 2?
(b) For what value of â€˜mâ€™ is x^{3} â€“ 2mx^{2} + 16 is divisible by (x + 2)?
Solution: (a) Here p(x) = x^{2 }+ 4x â€“ kx + 2
If p(x) is exactly divisible by x â€“ 2, then p(2) = 0
i.e. (2)^{2} + 4(2) â€“ k(2) + 2 = 0
â‡’ 4 + 8 â€“ 2k + 2 = 0
â‡’ 14 â€“ 2k = 0
â‡’ 2k = 14
â‡’ k= 14/2 = 7
Thus, the required value of k is 7.
(b) Here, p(x) = x^{3} â€“ 2mx^{2} + 16
âˆ´ p(â€“2) = (â€“2)^{3} â€“2(â€“2)^{2}m + 16
= â€“8 â€“8m + 16
= â€“8m + 8
Since, p(x) is divisible by x + 2
âˆ´ p(â€“2) = 0
or â€“8m + 8 = 0
â‡’ m = 1
Question 5. Factorise x^{2} â€“ x â€“ 12.
Solution: We have x^{2} â€“ x â€“ 12 = x^{2 }â€“ 4x + 3x â€“ 12
= x(x â€“ 4) + 3(x â€“ 4)
= (x â€“ 4)(x + 3)
Thus, x^{2 }â€“ x â€“ 12 = (x â€“ 4)(x + 3)
Question 6. If x + (1/2)x = 5, then find the value of x^{2} +
Solution: We have x + 1/2)x = 5
Squaring both sides, we get
â‡’
â‡’
â‡’
Thus, the requiewd value of
Question 1. Check whether (x â€“ 1) is a factor of the polynomial x^{3} â€“ 27x^{2 }+ 8x + 18.
Solution: Here, p(x) = x^{3 }â€“ 27x^{2} + 8x + 18 (x â€“ 1) will be a factor of p(x) only if (x â€“ 1) divides p(x) leaving a remainder 0.
For x â€“ 1 = 0
â‡’ x = 1
âˆ´ p(1) = (1)^{3} â€“ 27(1)^{2} + 8(1) + 18
= 1 â€“ 27 + 8 + 18
= 27 â€“ 27
= 0
Since, p(1) = 0
âˆ´ (x â€“ 1) is a factor of p(x).
Thus, (x â€“ 1) is a factor of x^{3} â€“ 27x^{2} + 8x + 18.
Question 2. Find the remainder when f(x) =
Solution: Here f(x) =
Divisor = x+(2/3)
since, x+(2/3) = 0 â‡’ x = -2/3
âˆ´ Remainder = f (-2/3)
i.e.
Remainder
Thus, the required remainder = (10/27)
Question 3. Find the value of k, if (x â€“ k) is a factor of x^{6} â€“ kx^{5} + x^{4} â€“ kx^{3} + 3x â€“ k + 4.
Solution: Here p(x) = x^{6} â€“ kx^{5} + x^{4 }â€“ kx^{3} + 3x â€“ k + 4
If (x â€“ k) is a factor of p(x), then p(k) = 0
i.e (k)^{6} â€“ k(k^{5}) + k^{4} â€“ k(k^{3}) + 3k â€“ k + 4 = 0
or k^{6 }â€“ k^{6} + k^{4} â€“ k^{4} + 3k â€“ k + 4 = 0
or 2k + 4 = 0
or 2k = â€“4
or k = (-4/2)
= â€“2
Thus, the required value of k is â€“2.
Question 4. Factorise: 9a^{2} â€“ 9b^{2 }+ 6a + 1
Solution: 9a^{2} â€“ 9b^{2} + 6a + 1
= [9a^{2} + 6a + 1] â€“ 9b^{2 }
= [(3a)^{2} + 2(3a)(1) + (1)2] â€“ (3b)^{2 }
= (3a + 1)^{2} â€“ (3b)^{2}
= [(3a + 1) + 3b][(3a + 1) â€“ 3b] [using x^{2 }â€“ y^{2} = (x â€“ y)(x + y)
= (3a + 1 + 3b)(3a + 1 â€“ 3b)
Question 5. Find the value of x3 + y3 â€“ 12xy + 64, when x + y = â€“4.
Solution: x^{3} + y^{3 }â€“ 12xy + 64
= (x)^{3} + (y)^{3} + (4)^{3 }â€“ 3(x)(y)(4)
= [x^{2} + y^{2} + 42 â€“ xy â€“ y. 4 â€“ 4 . x](x + y + 4)
= [x^{2} + y^{2} + 16 â€“ xy â€“ 4y â€“ 4x][x + y + 4] ...(1)
Since, x + y = â€“4 âˆ´ x + y + 4 = 0 ...(2)
From (1) and (2), we have x^{3} + y^{3 }â€“ 12xy + 64
= [x^{2 }+ y^{2} + 16 â€“ xy â€“ 4y â€“ 4x][0] = 0
Thus, x^{3} + y^{3} â€“ 12xy + 64 = 0.