Polynomials — Practice

Question 1

Which of the following is a polynomial: (a) √x + 3, (b) 1/x + 5, (c) 2x³ − 7x + 4, (d) sin(x) + 1?

Accepted Answer

Step 1 — Check each candidate against the polynomial rule.
  A polynomial requires every term to be a (constant) × x^k with k a non-negative integer.

Step 2 — Apply.
  (a) √x = x^(1/2). Fractional exponent → NOT a polynomial.
  (b) 1/x = x^(−1). Negative exponent → NOT a polynomial.
  (c) 2x³, −7x, 4 all have non-negative integer exponents → polynomial.
  (d) sin(x) is a transcendental function → NOT a polynomial.

✦ Answer: Only (c) is a polynomial.

Degree of (c) = 3 (highest power).

Question 2

Find the degree of the polynomial 4x⁵ − 7x² + 6.

Accepted Answer

Step 1 — Identify the highest power of x present.
  Terms: 4x⁵, −7x², 6. The highest power is 5.

✦ Answer: Degree = 5.

Rule: degree = highest exponent of the variable, regardless of the coefficient's sign or value.

Question 3

Find p(3) if p(x) = 2x² − 5x + 1.

Accepted Answer

Step 1 — Substitute x = 3.
  p(3) = 2·(3)² − 5·(3) + 1.

Step 2 — Evaluate.
  = 2·9 − 15 + 1
  = 18 − 15 + 1
  = 4.

✦ Answer: p(3) = 4.

Question 4

Is x = −2 a zero of p(x) = x³ + 2x² − 5x − 6?

Accepted Answer

Step 1 — Definition of zero: a is a zero iff p(a) = 0.

Step 2 — Compute p(−2).
  p(−2) = (−2)³ + 2·(−2)² − 5·(−2) − 6
        = −8 + 2·4 + 10 − 6
        = −8 + 8 + 10 − 6
        = 4.

Step 3 — Conclude.
  p(−2) = 4 ≠ 0.

✦ Answer: NO, x = −2 is not a zero (p(−2) = 4).

Question 5

Expand (3x + 4)² using an identity.

Accepted Answer

Step 1 — Use identity (a + b)² = a² + 2ab + b² with a = 3x, b = 4.

Step 2 — Substitute.
  (3x)² + 2·(3x)·(4) + (4)²
  = 9x² + 24x + 16.

✦ Answer: 9x² + 24x + 16.

Memo: square the first term, square the second, sum cross-product doubled.

Question 6

Without computing, find 102 × 98 using an identity.

Accepted Answer

Step 1 — Notice the pattern: 102 = 100 + 2 and 98 = 100 − 2.
  This is exactly the (a + b)(a − b) form with a = 100, b = 2.

Step 2 — Apply identity 3: (a + b)(a − b) = a² − b².
  102 × 98 = 100² − 2² = 10000 − 4 = 9996.

✦ Answer: 102 × 98 = 9996.

Trick: any 'nearly-equal' product can be done mentally with this identity.

Question 7

Find the remainder when p(x) = x³ + 3x² − 5x + 7 is divided by (x − 2).

Accepted Answer

Step 1 — Remainder Theorem: p(x) ÷ (x − a) gives remainder p(a).
  Here a = 2.

Step 2 — Compute p(2).
  p(2) = (2)³ + 3·(2)² − 5·(2) + 7
       = 8 + 12 − 10 + 7
       = 17.

✦ Answer: Remainder = 17.

Far faster than long division. Always use the theorem when the divisor is linear.

Question 8

Find the remainder when p(x) = 3x³ − 2x² + 4x − 1 is divided by (3x − 1).

Accepted Answer

Step 1 — Find the root of the divisor.
  3x − 1 = 0 ⇒ x = 1/3.
  By the general Remainder Theorem, the remainder is p(1/3).

Step 2 — Compute p(1/3).
  p(1/3) = 3·(1/27) − 2·(1/9) + 4·(1/3) − 1
         = 1/9 − 2/9 + 4/3 − 1.

Step 3 — Combine over a common denominator of 9.
         = 1/9 − 2/9 + 12/9 − 9/9
         = (1 − 2 + 12 − 9)/9
         = 2/9.

✦ Answer: Remainder = 2/9.

Working with fractions feels slow but is much faster than long-dividing a cubic by (3x − 1).

Question 9

Determine whether (x + 1) is a factor of p(x) = x⁴ − x³ − 2x + 2.

Accepted Answer

Step 1 — Apply the Factor Theorem.
  (x + 1) is a factor of p(x) iff p(−1) = 0.

Step 2 — Compute p(−1).
  p(−1) = (−1)⁴ − (−1)³ − 2·(−1) + 2
        = 1 − (−1) + 2 + 2
        = 1 + 1 + 2 + 2
        = 6.

Step 3 — Conclude.
  p(−1) = 6 ≠ 0.

✦ Answer: NO, (x + 1) is NOT a factor (p(−1) = 6).

Question 10

Factorise 6x² + 11x − 10 by splitting the middle term.

Accepted Answer

Step 1 — Compute a·c (the product of leading coefficient and constant term).
  a·c = 6 · (−10) = −60.

Step 2 — Find two numbers with product −60 and sum 11 (the middle coefficient).
  Factors of 60: (1, 60), (2, 30), (3, 20), (4, 15), (5, 12), (6, 10).
  With opposite signs to give a negative product:
  15 and −4 → 15 × (−4) = −60 ✓, 15 + (−4) = 11 ✓.

Step 3 — Split the middle term.
  6x² + 15x − 4x − 10.

Step 4 — Group and factor.
  3x(2x + 5) − 2(2x + 5) = (3x − 2)(2x + 5).

Step 5 — Verify by expansion.
  (3x − 2)(2x + 5) = 6x² + 15x − 4x − 10 = 6x² + 11x − 10. ✓

✦ Answer: 6x² + 11x − 10 = (3x − 2)(2x + 5).

Question 11

Factorise 25x² − 49y² using an identity.

Accepted Answer

Step 1 — Recognise the difference-of-squares pattern.
  25x² = (5x)², 49y² = (7y)². So the expression is a² − b² with a = 5x, b = 7y.

Step 2 — Apply identity 3.
  a² − b² = (a + b)(a − b) = (5x + 7y)(5x − 7y).

✦ Answer: 25x² − 49y² = (5x + 7y)(5x − 7y).

Question 12

Factorise 8x³ + 27.

Accepted Answer

Step 1 — Recognise as a sum of cubes.
  8x³ = (2x)³, 27 = 3³. So this is a³ + b³ with a = 2x, b = 3.

Step 2 — Apply identity a³ + b³ = (a + b)(a² − ab + b²).
  = (2x + 3)((2x)² − (2x)(3) + 3²)
  = (2x + 3)(4x² − 6x + 9).

Step 3 — Check if the quadratic factor factorises further.
  Discriminant of 4x² − 6x + 9 = (−6)² − 4·4·9 = 36 − 144 = −108 < 0.
  No real roots ⇒ irreducible over ℝ.

✦ Answer: 8x³ + 27 = (2x + 3)(4x² − 6x + 9).

Question 13

Expand (2x − 3y)³.

Accepted Answer

Step 1 — Apply identity (a − b)³ = a³ − 3a²b + 3ab² − b³.
  Set a = 2x, b = 3y.

Step 2 — Substitute.
  = (2x)³ − 3·(2x)²·(3y) + 3·(2x)·(3y)² − (3y)³
  = 8x³ − 3·4x²·3y + 3·2x·9y² − 27y³
  = 8x³ − 36x²y + 54xy² − 27y³.

✦ Answer: 8x³ − 36x²y + 54xy² − 27y³.

Check by counting terms: a cube of a binomial always has 4 terms — correct here.

Question 14

Factorise x³ − 6x² + 11x − 6 completely.

Accepted Answer

Step 1 — Find one zero using the Factor Theorem.
  Try x = 1: p(1) = 1 − 6 + 11 − 6 = 0. ✓
  So (x − 1) is a factor.

Step 2 — Divide p(x) by (x − 1).
  Long division (or synthetic):
  x³ − 6x² + 11x − 6 ÷ (x − 1) = x² − 5x + 6.

Step 3 — Factorise the resulting quadratic.
  x² − 5x + 6: need two numbers with product 6 and sum −5.
  Numbers: −2 and −3.
  x² − 2x − 3x + 6 = x(x − 2) − 3(x − 2) = (x − 2)(x − 3).

Step 4 — Combine.
  p(x) = (x − 1)(x − 2)(x − 3).

Step 5 — Verify by checking each zero.
  p(1) = 0 ✓, p(2) = 8 − 24 + 22 − 6 = 0 ✓, p(3) = 27 − 54 + 33 − 6 = 0 ✓.

✦ Answer: x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3).

Zeros: 1, 2, 3.

Question 15

If a + b + c = 0, find the value of a³ + b³ + c³ − 3abc.

Accepted Answer

Step 1 — Recall the three-variable identity.
  a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Step 2 — Substitute the given condition.
  a + b + c = 0 ⇒ the entire RHS is 0 × (anything) = 0.

Step 3 — Therefore a³ + b³ + c³ − 3abc = 0, i.e. a³ + b³ + c³ = 3abc.

✦ Answer: a³ + b³ + c³ − 3abc = 0 (equivalently, a³ + b³ + c³ = 3abc).

Follow-up: Find 5³ + (−2)³ + (−3)³. Since 5 + (−2) + (−3) = 0, the trick says 5³ + (−2)³ + (−3)³ = 3·5·(−2)·(−3) = 90. Verify: 125 − 8 − 27 = 90. ✓

Question 16

If a + b = 5 and ab = 6, find a³ + b³ without solving for a, b.

Accepted Answer

Step 1 — Use the identity (a + b)³ = a³ + b³ + 3ab(a + b).

Step 2 — Rearrange to isolate a³ + b³.
  a³ + b³ = (a + b)³ − 3ab(a + b).

Step 3 — Substitute the given values.
  = 5³ − 3·6·5
  = 125 − 90
  = 35.

✦ Answer: a³ + b³ = 35.

Why elegant: we never had to solve the quadratic x² − 5x + 6 = 0 (whose roots are a, b = 2, 3). Always use symmetric identities when only sum and product are given.

Question 17

If (x − 2) and (x + 1) are both factors of p(x) = x³ + ax² + bx − 6, find a and b.

Accepted Answer

Step 1 — Apply the Factor Theorem twice.
  (x − 2) is a factor ⇒ p(2) = 0.
  (x + 1) is a factor ⇒ p(−1) = 0.

Step 2 — Write both conditions.
  p(2) = 8 + 4a + 2b − 6 = 0 ⇒ 4a + 2b = −2 ⇒ 2a + b = −1. … (i)
  p(−1) = −1 + a − b − 6 = 0 ⇒ a − b = 7. … (ii)

Step 3 — Solve the simultaneous system.
  Add (i) and (ii): (2a + b) + (a − b) = −1 + 7 ⇒ 3a = 6 ⇒ a = 2.
  Substitute in (ii): 2 − b = 7 ⇒ b = −5.

Step 4 — Verify.
  p(x) = x³ + 2x² − 5x − 6.
  p(2) = 8 + 8 − 10 − 6 = 0 ✓.
  p(−1) = −1 + 2 + 5 − 6 = 0 ✓.

✦ Answer: a = 2, b = −5.

Question 18

Find all integer zeros of p(x) = x⁴ − 5x² + 4.

Accepted Answer

Step 1 — Notice this is a quadratic in x² — a 'biquadratic'.
  Let y = x². Then p becomes q(y) = y² − 5y + 4.

Step 2 — Factorise the quadratic in y.
  q(y) = (y − 1)(y − 4).

Step 3 — Substitute back y = x².
  p(x) = (x² − 1)(x² − 4).

Step 4 — Apply the difference-of-squares identity twice.
  x² − 1 = (x − 1)(x + 1).
  x² − 4 = (x − 2)(x + 2).

Step 5 — Complete factorisation.
  p(x) = (x − 1)(x + 1)(x − 2)(x + 2).

Step 6 — Read off the zeros.
  Zeros: 1, −1, 2, −2.

✦ Answer: Four integer zeros — namely ±1 and ±2.

Practice — Polynomials

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