RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Maths · 11 questions · 26 marks
A polynomial is just a tidy sum of terms like 4x³ - 7x + 2, and almost every algebraic technique you meet later rests on handling them well. This chapter teaches you to classify them by degree, evaluate them at a value, and split them into factors using the Remainder and Factor Theorems. Once you can spot a zero by inspection, factorising a cubic stops feeling like guesswork.
Which of the following expressions is NOT a polynomial in x?
Answer
4x^(1/2) + x - 6 is not a polynomial. In a polynomial every exponent of the variable must be a whole number. Here x^(1/2) = √x has exponent 1/2, which is not a whole number, so the expression fails the definition. The other three are valid: 7x³ - 2x + 9 is cubic, 5x² + (3/2)x is quadratic (a fractional coefficient is allowed, only fractional exponents are not), and 11 is a constant polynomial of degree 0.
When p(x) = x³ - 4x² + 5x - 2 is divided by (x - 3), the remainder is:
Answer
The remainder is 4. By the Remainder Theorem the remainder equals p(3). Substituting: p(3) = 27 - 4(9) + 5(3) - 2 = 27 - 36 + 15 - 2 = 4. Since the remainder is not zero, (x - 3) is not a factor of p(x).
The zero of the linear polynomial 6x + 15 is:
Answer
The zero is -5/2. A zero is the value of x that makes the polynomial equal 0. Set 6x + 15 = 0, so 6x = -15 and x = -15/6, which simplifies to -5/2. The option 5/2 is wrong in sign, and -2/5 comes from accidentally inverting the fraction -b/a instead of computing it.
If x + y + z = 0, then x³ + y³ + z³ is equal to:
Answer
It equals 3xyz. The general identity is x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx). When x + y + z = 0 the entire right-hand side becomes 0, so x³ + y³ + z³ - 3xyz = 0, giving x³ + y³ + z³ = 3xyz.
Assertion (A): (x + 4) is a factor of p(x) = x³ + 3x² - 6x + 8. Reason (R): By the Factor Theorem, (x - a) is a factor of p(x) if and only if p(a) = 0.
Answer
A is false but R is true. Reason R is the correct statement of the Factor Theorem. Testing the Assertion, (x + 4) is a factor only if p(-4) = 0. Here p(-4) = (-64) + 3(16) - 6(-4) + 8 = -64 + 48 + 24 + 8 = 16, which is not zero. So (x + 4) is not a factor and Assertion A is false.
Find the value of k for which (x - 2) is a factor of p(x) = 3x³ - kx² + 2x - 8.
Answer
By the Factor Theorem, (x - 2) is a factor only if p(2) = 0. p(2) = 3(8) - k(4) + 2(2) - 8 = 24 - 4k + 4 - 8 = 20 - 4k. Setting 20 - 4k = 0 gives 4k = 20, so k = 5. Hence (x - 2) is a factor when k = 5.
Factorise 6x² + 11x - 35 by splitting the middle term.
Answer
Here a = 6, b = 11 and c = -35, so ac = 6 × (-35) = -210. We need two numbers whose product is -210 and whose sum is +11. These are +21 and -10. Split the middle term: 6x² + 21x - 10x - 35. Group: 3x(2x + 7) - 5(2x + 7). Take out the common factor: (2x + 7)(3x - 5). So 6x² + 11x - 35 = (2x + 7)(3x - 5).
Without actually dividing, find the remainder when p(x) = 2x⁴ - 5x² + 7 is divided by (x + 2).
Answer
By the Remainder Theorem, dividing by (x + 2) = (x - (-2)) gives remainder p(-2). p(-2) = 2(-2)⁴ - 5(-2)² + 7 = 2(16) - 5(4) + 7 = 32 - 20 + 7 = 19. The remainder is 19.
Factorise the cubic polynomial p(x) = x³ - 2x² - 13x - 10 completely, showing how you locate the first factor.
Answer
Step 1 — Search for a zero among the factors of the constant term -10, i.e. ±1, ±2, ±5, ±10. p(-1) = -1 - 2 + 13 - 10 = 0. So x = -1 is a zero and, by the Factor Theorem, (x + 1) is a factor. Step 2 — Divide p(x) by (x + 1). The quotient is x² - 3x - 10, since (x + 1)(x² - 3x - 10) expands back to x³ - 2x² - 13x - 10. Step 3 — Factorise the quadratic by splitting the middle term: we need two numbers with product -10 and sum -3, namely -5 and +2. x² - 5x + 2x - 10 = x(x - 5) + 2(x - 5) = (x - 5)(x + 2). Step 4 — Combine: p(x) = (x + 1)(x + 2)(x - 5). The three zeros are therefore x = -1, x = -2 and x = 5.
(a) If p(x) = x³ + ax² + bx - 12 leaves remainder 0 when divided by (x - 2) and remainder -18 when divided by (x + 1), find a and b. (b) Using the values found, evaluate p(1).
Answer
(a) By the Remainder Theorem, p(2) = 0 and p(-1) = -18. p(2) = 8 + 4a + 2b - 12 = 4a + 2b - 4. Setting this to 0 gives 4a + 2b = 4, i.e. 2a + b = 2. ...(i) p(-1) = -1 + a - b - 12 = a - b - 13. Setting this to -18 gives a - b = -5. ...(ii) Add (i) and (ii): (2a + b) + (a - b) = 2 + (-5), so 3a = -3 and a = -1. Substitute in (ii): -1 - b = -5, so b = 4. Hence a = -1 and b = 4. (b) With a = -1 and b = 4, p(x) = x³ - x² + 4x - 12. p(1) = 1 - 1 + 4 - 12 = -8.
Read the following and answer the questions that follow: A carpenter builds an open rectangular storage box from a sheet of plywood. The height of the box is x cm, the breadth is (x + 2) cm and the length is (x + 5) cm, so the volume in cubic centimetres is given by the polynomial V(x) = x(x + 2)(x + 5). (a) Write V(x) in expanded standard form and state its degree. (b) Find the volume when the height is 4 cm. (c) Show that x = -2 is a zero of V(x), and explain why this zero has no physical meaning for the box. (d) If the carpenter needs a volume of 180 cm³, verify whether x = 4 is a suitable height.
Answer
(a) V(x) = x(x + 2)(x + 5). First (x + 2)(x + 5) = x² + 7x + 10, so V(x) = x³ + 7x² + 10x. The highest power is 3, so the degree is 3 (a cubic polynomial). (b) V(4) = 4³ + 7(4²) + 10(4) = 64 + 112 + 40 = 216. The volume is 216 cm³. (c) V(-2) = (-2)³ + 7(-2)² + 10(-2) = -8 + 28 - 20 = 0, so x = -2 is indeed a zero. It has no physical meaning because x represents the height of a real box, and a length can never be negative; only positive values of x are admissible here. (d) From part (b), V(4) = 216 cm³, not 180 cm³. So x = 4 cm does not give the required volume — the required height must be a little less than 4 cm, since V(3) = 27 + 63 + 30 = 120 cm³ and the volume increases with x between 3 and 4.
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