Polynomials
Class 9 Maths · Polynomials · 33 questions with answers
Polynomials is the first chapter of Class 9 and it assumes Class 8 completely - expanding a bracket and factorising an expression have to be automatic before any of this works. Everything here rests on one small idea: to find the remainder when dividing by (x - a), substitute x = a. That single substitution replaces long division and unlocks the whole chapter.
- States the degree and finds the value of a polynomial at a point
- Uses the remainder theorem instead of long division
- Uses the factor theorem to test a factor and to find an unknown coefficient
- Factorises a cubic completely
- Applies the expanded identities for three terms and for cubes
Exercise 1 of 3 · 11 questions
Answer each one.
Write the answer · Warm-up
Here’s one done for you
p(2) = 0 for p(x) = x² - 3x + 2, which means 2 is a zero of the polynomial. Notice what that also tells you: since p(2) = 0, the factor theorem says (x - 2) is a factor. Every 'find p(a)' question is quietly a factor question, and seeing that early makes the rest of the chapter much shorter.
- 1)What is the degree of p(x) = 5x³ - 2x + 7?
- 2)If p(x) = x² - 3x + 2, find p(2).
- 3)If p(x) = x³ + 2x² - x + 1, find p(-1).
- 4)Is x = 3 a zero of p(x) = x² - 5x + 6?
- 5)Find the zero of p(x) = 2x + 5.
- 6)What is the degree of p(x) = 7 - 2x + x⁴?
- 7)If p(x) = 2x² + x - 6, find p(-1).
- 8)If p(x) = x³ - 4x, find p(0) and p(2).
- 9)Is x = -2 a zero of p(x) = x² + 5x + 6?
- 10)Find the zero of p(x) = 3x - 7.
- 11)Write the coefficient of x² in p(x) = 5 - 3x² + x³.
Answers
- 1) 3 - the highest power of x present.
- 2) p(2) = 4 - 6 + 2 = 0. So 2 is a zero of this polynomial.
- 3) p(-1) = -1 + 2 + 1 + 1 = 3.
- 4) Yes. p(3) = 9 - 15 + 6 = 0, and a zero is a value that makes the polynomial zero.
- 5) x = -5/2, from 2x + 5 = 0.
- 6) 4. The degree is the highest power of x, whatever order the terms are written in.
- 7) p(-1) = 2(1) + (-1) - 6 = -5.
- 8) p(0) = 0 and p(2) = 8 - 8 = 0, so 0 and 2 are both zeroes.
- 9) Yes. p(-2) = 4 - 10 + 6 = 0, and a zero is a value that makes p(x) = 0.
- 10) 3x - 7 = 0, so x = 7/3.
- 11) -3. The coefficient carries its sign.
Exercise 2 of 3 · 11 questions
Use the remainder or factor theorem. Do not use long division.
Write the answer · Guided
Here’s one done for you
For the remainder on dividing by (2x - 1), set 2x - 1 = 0 rather than reaching for x = 1. That gives x = 1/2, and substituting gives 3/2. Children who have learned the theorem as 'put x = a for (x - a)' get this one wrong every time. The rule is really 'substitute the value that makes the divisor zero', which covers every case including this one.
- 1)Find the remainder when x³ - 3x² + 4x - 5 is divided by (x - 2).
- 2)Find the remainder when x³ + 3x² + 3x + 1 is divided by (x + 1).
- 3)Is (x - 1) a factor of x³ - 2x² + x?
- 4)Find k if (x - 2) is a factor of x³ + kx² + 2x - 4.
- 5)Find the remainder when 4x³ - 12x² + 14x - 3 is divided by (2x - 1).
- 6)Find the remainder when x³ + x² - 2x + 1 is divided by (x - 2).
- 7)Find the remainder when 2x³ - 3x² + 4x - 5 is divided by (x - 1).
- 8)Is (x - 1) a factor of x³ - 6x² + 11x - 6?
- 9)Find k if (x + 1) is a factor of x³ + 3x² + kx + 4.
- 10)Find the remainder when x⁴ - x² + 2 is divided by (x + 1).
- 11)Show that (x - 3) is a factor of x³ - 6x² + 11x - 6.
Answers
- 1) p(2) = 8 - 12 + 8 - 5 = -1. The remainder is -1.
- 2) p(-1) = -1 + 3 - 3 + 1 = 0. The remainder is 0, so (x + 1) is a factor.
- 3) Yes. p(1) = 1 - 2 + 1 = 0, so by the factor theorem (x - 1) is a factor.
- 4) p(2) = 8 + 4k + 4 - 4 = 8 + 4k. For (x - 2) to be a factor this must be zero, so k = -2.
- 5) Set 2x - 1 = 0, giving x = 1/2. Then p(1/2) = 1/2 - 3 + 7 - 3 = 3/2. The remainder is 3/2.
- 6) p(2) = 8 + 4 - 4 + 1 = 9. The remainder is 9.
- 7) p(1) = 2 - 3 + 4 - 5 = -2. The remainder is -2.
- 8) Yes. p(1) = 1 - 6 + 11 - 6 = 0, so by the factor theorem (x - 1) is a factor.
- 9) p(-1) = -1 + 3 - k + 4 = 0, so 6 - k = 0 and k = 6.
- 10) p(-1) = 1 - 1 + 2 = 2. The remainder is 2.
- 11) p(3) = 27 - 54 + 33 - 6 = 0, so by the factor theorem (x - 3) is a factor.
Exercise 3 of 3 · 11 questions
Factorise completely. Find one factor by the factor theorem first.
Write the answer · Practice
Here’s one done for you
For x³ - 3x² - 9x - 5, try small values first: p(1) = -16, p(-1) = -1 - 3 + 9 - 5 = 0. So (x + 1) is a factor. Dividing gives x² - 4x - 5, which factorises to (x + 1)(x - 5), so the full answer is (x + 1)²(x - 5). Two habits make cubics reliable: try ±1 and ±2 before anything else, since exam cubics almost always have a small integer zero, and check the answer by expanding it back.
- 1)x³ - 2x² - x + 2
- 2)2y³ + y² - 2y - 1
- 3)x³ - 3x² - 9x - 5
- 4)x³ - 23x² + 142x - 120
- 5)Expand (2x + 3y)³.
- 6)x³ - 6x² + 11x - 6
- 7)x³ + 6x² + 11x + 6
- 8)Expand (x + 2y)³.
- 9)Expand (3a - b)³.
- 10)Factorise 8x³ + 27y³.
- 11)Factorise x³ - 64.
Answers
- 1) (x - 1)(x - 2)(x + 1). Since p(1) = 0, divide by (x - 1) to get x² - x - 2, which factorises further.
- 2) (y - 1)(y + 1)(2y + 1). Since p(1) = 0, divide by (y - 1) to get 2y² + 3y + 1.
- 3) (x + 1)²(x - 5). Since p(-1) = 0, divide by (x + 1) to get x² - 4x - 5 = (x + 1)(x - 5).
- 4) (x - 1)(x - 10)(x - 12). Since p(1) = 0, divide by (x - 1) to get x² - 22x + 120.
- 5) 8x³ + 36x²y + 54xy² + 27y³, using (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = 2x and b = 3y.
- 6) (x - 1)(x - 2)(x - 3). p(1) = 0 gives the first factor, and dividing leaves x² - 5x + 6.
- 7) (x + 1)(x + 2)(x + 3). p(-1) = 0 gives the first factor.
- 8) x³ + 6x²y + 12xy² + 8y³.
- 9) 27a³ - 27a²b + 9ab² - b³.
- 10) (2x + 3y)(4x² - 6xy + 9y²), using a³ + b³ = (a + b)(a² - ab + b²).
- 11) (x - 4)(x² + 4x + 16), using a³ - b³ = (a - b)(a² + ab + b²).
While your child works
- The rule is 'substitute the value that makes the divisor zero', not 'put x = a'. That covers divisors like (2x - 1), which catch out children who learned the shortcut.
- For cubics, try ±1 and ±2 first. Exam questions are built to have a small integer zero.
- Always expand the factorised answer back. Fifteen seconds, and it catches every error.