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Value of a polynomial · 3 marks

Find the value of the quadratic polynomial 7s2 − 4s + 6 if:

  1. (i) s = 0
  2. (ii) s = −3
  3. (iii) s = 4
Answer: (i) 6 (ii) 81 (iii) 102

Step-by-step solution

Idea: Substitute the value of s in every term. Square first, then multiply by 7; a negative number squared is positive.

(i) s = 0

  1. 7(0)2 − 4(0) + 6½ mark
  2. = 0 − 0 + 6 = 6½ mark
6

(ii) s = −3

  1. 7(−3)2 − 4(−3) + 6 = 7 × 9 + 12 + 6½ mark
  2. = 63 + 12 + 6 = 81½ mark
81

(iii) s = 4

  1. 7(4)2 − 4(4) + 6 = 7 × 16 − 16 + 6½ mark
  2. = 112 − 16 + 6 = 102½ mark
102
(i) 6 (ii) 81 (iii) 102

Check: At s = 0 only the constant term is left, so the value must be 6 ✓. For (ii), (−3)2 = +9 and −4 × (−3) = +12, so all three terms are positive: 63 + 12 + 6 = 81 ✓.

Answer to write in the exam

(i)

7(0)2 − 4(0) + 6 = 0 − 0 + 6

∴ Value = 6

(ii)

7(−3)2 − 4(−3) + 6 = 63 + 12 + 6

∴ Value = 81

(iii)

7(4)2 − 4(4) + 6 = 112 − 16 + 6

∴ Value = 102

Common mistakes that cost marks

  • Writing (−3)2 = −9. A negative number times itself is positive: (−3) × (−3) = 9.
  • Working out 7s2 as (7s)2, for example (28)2 at s = 4. Only s is squared: 7 × 16 = 112.
  • Writing −4(−3) = −12. Minus times minus is plus: +12.

How this can come in the exam

MCQ (1 mark)

The value of 2t2 + 3t − 5 at t = −2 is

  1. −3
  2. 9
  3. −19
  4. −7
Show answer

(A) −3
2(4) + 3(−2) − 5 = 8 − 6 − 5 = −3.

Assertion–Reason (1 mark)

Assertion (A): For q(s) = s2 + 1, q(−5) = q(5).
Reason (R): A polynomial always takes the same value at s and −s.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Show answer

(C) A is true but R is false.
A is true: both equal 26. R is false: for example, 5s − 3 gives 22 at s = 5 and −28 at s = −5.

Try one yourself

Find the value of 3y2 − 5y + 2 at y = −1 and at y = 3.

Show answer

At −1: 3 + 5 + 2 = 10. At 3: 27 − 15 + 2 = 14.

More questions like this

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