CLASS IX
MATHEMATICS WORKSHEET - ANSWER KEY
CHAPTER 2: POLYNOMIALS
Very Short Answer Type Questions
Q1. An example of an algebraic expression that is not a polynomial is √(x^3 + 1). This is because the expression contains a radical, and radicals are not allowed in polynomials.
Q2. The expression p(x) = √(x^3 + 1) is not a polynomial because it contains a radical. Radicals are not allowed in polynomials because they are not defined for all real numbers. For example, the radical √(-1) is undefined.
Q3. The value of the polynomial 8x^3 - 6x^2 + 2 at x = 1 is 2.
Q4. The value of the polynomial p(x) = 6x^3 + 5x^2 – 3x + 2 at x = -1 is -19.
Q5. The zero(s) of the polynomial p(y) = 2y + 7 are -7/2 and 0.
Q6. The remainder when x^101 – 1 is divided by x - 1 is 100.
Q7. The polynomial xn + yn is not divisible by x - y (y ≠ 0) in general. For example, if n = 2, then x^2 + y^2 = (x - y)(x + y) + y^2, so x^2 + y^2 is not divisible by x - y.
Q8. The polynomials in standard form are:
i. 4y - 4y^3 + 3 - y^4 = -y^4 + 4y^3 - 4y + 3
ii. 5m^3 - 2m^2 - 6m + 7 = 5m^3 - 2m^2 - 6m + 7
Short Answer Type Questions
Q10. If y = -1 is a zero of the polynomial q(y) = 4y^3 + ky^2 - y - 1, then k = -11.
Q11. The value of m for which x^3 – 2mx^2 + 16 is divisible by x + 2 is 4.
Q12. The identity (a + b + c)^3 – a^3 – b^3 – c^3 = 3(a + b)(b + c)(c + a) can be proved by expanding the left-hand side and using the distributive property.
Q13. If x + 1/x = 5, then x^3 + 1/x^3 = 125.
Q14. The value of a in each of the cases is:
i. R1 = R2 --> a = -10
ii. R1 + R2 = 0 --> a = 6
iii. 2R1 + R2 = 0 --> a = -6
Q15. The value of a^2 + b^2 + c^2 is 85.
Q16. The identity a^2bc + ab^2c + abc^2 = 3abc can be proved by expanding the left-hand side and using the distributive property.
Q17. The zeroes of the polynomial (x - 2)^2 – (x + 2)^2 are 2 and -2.
To find the zeroes of the polynomial, we can factor it as follows:
(x - 2)^2 – (x + 2)^2 = (x - 2 + x + 2)(x - 2 - x - 2) = 2(-4) = -8
The polynomial is equal to 0 when x = 2 or x = -2, so the zeroes of the polynomial are 2 and -2.
Q18. The polynomial p(x) = x^4 + x^3 – 7x^2 – x + 6 can be factorized using the factor theorem. The factor theorem states that if a is a zero of the polynomial p(x), then x - a is a factor of p(x). In this case, we can see that x = 1 is a zero of p(x), so x - 1 is a factor of p(x).
We can factor p(x) as follows:
p(x) = (x - 1)(x^3 + 2x^2 + 7x + 6)
Long Answer Type Questions
Q18. The polynomial p(x) = x^4 + x^3 – 7x^2 – x + 6 can be factorized using the factor theorem. The factor theorem states that if a is a zero of the polynomial p(x), then x - a is a factor of p(x). In this case, we can see that x = 1 is a zero of p(x), so x - 1 is a factor of p(x).
We can factor p(x) as follows:
p(x) = (x - 1)(x^3 + 2x^2 + 7x + 6)
Q19. The polynomial 2x^4 – 6x^3 + 3x^2 + 3x – 2 is exactly divisible by x^2 – 3x + 2. This can be proved by using the polynomial remainder theorem. The polynomial remainder theorem states that if a polynomial p(x) is divided by x - a, then the remainder is p(a).
In this case, if we divide 2x^4 – 6x^3 + 3x^2 + 3x – 2 by x^2 – 3x + 2, then the remainder is 2. This means that 2x^4 – 6x^3 + 3x^2 + 3x – 2 is exactly divisible by x^2 – 3x + 2.
Q20. The remainder when the polynomial p(x) = x^4 – 2x^3 + 3x^2 – ax + b is divided by x – 1 is 5, and the remainder when it is divided by x + 1 is 19. This means that the polynomial p(x) – 5 is divisible by x – 1, and the polynomial p(x) – 19 is divisible by x + 1.
We can use the polynomial remainder theorem to find the values of a and b. The polynomial remainder theorem states that if a polynomial p(x) is divided by x - a, then the remainder is p(a).
In this case, the remainder when p(x) – 5 is divided by x – 1 is 0, so p(1) = 5. This means that a = 1.
The remainder when p(x) – 19 is divided by x + 1 is 0, so p(-1) = 19. This means that b = 24.
Therefore, the remainder when p(x) is divided by x – 2 is -12.
Q21. The expression (4x^2 - 9y^2)^3 + (9y^2 - 16y^2)^3 + (16z^2 - 4x^2)^3 / (2x - 3y)^3 + (3y - 4z)^3 + (4z - 2x)^3 can be simplified as follows:
(4x^2 - 9y^2)^3 + (9y^2 - 16y^2)^3 + (16z^2 - 4x^2)^3 / (2x - 3y)^3 + (3y - 4z)^3 + (4z - 2x)^3 =
3(4x^2 - 9y^2)(9y^2 - 16y^2)(16z^2 - 4x^2) / (2x - 3y)(3y - 4z)(4z - 2x) =
108x^2 y^2 z^2
Q22. If x – 3 and x – 1/3 are both factors of ax^2 + 5x + b, then a = b.
To prove this, we can use the factor theorem. The factor theorem states that if a is a zero of the polynomial p(x), then x - a is a factor of p(x).
In this case, x = 3 and x = 1/3 are both zeros of ax^2 + 5x + b, so (x - 3) and (x - 1/3) are both factors of ax^2 + 5x + b. This means that the polynomial ax^2 + 5x + b is divisible by both (x - 3) and (x - 1/3).
The only way that a polynomial can be divisible by both (x - 3) and (x - 1/3) is if it is a multiple of the least common multiple of (x - 3) and (x - 1/3). The least common multiple of (x - 3) and (x - 1/3) is 9x - 1, so ax^2 + 5x + b must be a multiple of 9x - 1.
Since ax^2 + 5x + b is a polynomial, it must be of the form 9x - 1 + c, where c is a constant. Therefore, a = 9 and b = 9 + c = c, so a = b.
Q23. The polynomials can be factorized as follows:
i. 3(x + 2)^2 – 5(x + 2) + 2 = (x + 2)(3x + 2)
ii. x^6 + y^6 = (x^3 + y^3)(x^3 – y^3) = (x + y)(x^2 – xy + y^2)(x – y)(x^2 + xy + y^2)
iii. 3√(3x^3) – 5√(5y^3) = 3√3x – 5√5y = 3x^(1/3) – 5y^(1/3)
