Find the zeroes of each quadratic polynomial below, and verify in each case that the sum of the zeroes is −b/a and the product is c/a: (i) x² − 2x − 8, (ii) 4s² − 4s + 1, (iii) 6x² − 3 − 7x, (iv) 4u² + 8u, (v) t² − 15, (vi) 3x² − x − 4.
Hint. Factorise by splitting the middle term, set each factor to zero, then compare your two zeroes against −b/a and c/a read straight off the polynomial.
Every part follows the same three moves: factorise, read off the zeroes, then check them against the coefficients.
(i) x² − 2x − 8. Here a = 1, b = −2, c = −8.
Split the middle term — you want two numbers multiplying to −8 and adding to −2: those are −4 and +2.
x² − 4x + 2x − 8 = x(x − 4) + 2(x − 4) = (x − 4)(x + 2), so the zeroes are 4 and −2.
Check: sum = 4 + (−2) = 2, and −b/a = 2 ✓. Product = 4 × (−2) = −8, and c/a = −8 ✓
(ii) 4s² − 4s + 1. Here a = 4, b = −4, c = 1.
This is a perfect square: 4s² − 4s + 1 = (2s − 1)². Setting 2s − 1 = 0 gives s = 1/2 — twice.
A repeated zero still counts twice when you check the relationships. Sum = 1/2 + 1/2 = 1, and −b/a = 4/4 = 1 ✓. Product = 1/2 × 1/2 = 1/4, and c/a = 1/4 ✓
(iii) 6x² − 3 − 7x. First put it in standard order: 6x² − 7x − 3, so a = 6, b = −7, c = −3.
Split −7x using two numbers with product 6 × (−3) = −18 and sum −7: those are −9 and +2.
6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (2x − 3)(3x + 1), giving zeroes 3/2 and −1/3.
Check: sum = 3/2 − 1/3 = 7/6, and −b/a = 7/6 ✓. Product = (3/2)(−1/3) = −1/2, and c/a = −3/6 = −1/2 ✓
(iv) 4u² + 8u. Here a = 4, b = 8, c = 0.
Do not reach for a formula — just take out the common factor: 4u² + 8u = 4u(u + 2). Zeroes are 0 and −2.
Check: sum = 0 + (−2) = −2, and −b/a = −8/4 = −2 ✓. Product = 0 × (−2) = 0, and c/a = 0/4 = 0 ✓
(v) t² − 15. Here a = 1, b = 0, c = −15.
A difference of two squares: t² − 15 = (t − √15)(t + √15), so the zeroes are √15 and −√15.
Check: sum = 0, and −b/a = 0 ✓. Product = (√15)(−√15) = −15, and c/a = −15 ✓
(vi) 3x² − x − 4. Here a = 3, b = −1, c = −4.
Split −x using two numbers with product 3 × (−4) = −12 and sum −1: those are −4 and +3.
3x² − 4x + 3x − 4 = x(3x − 4) + 1(3x − 4) = (3x − 4)(x + 1), giving zeroes 4/3 and −1.
Check: sum = 4/3 − 1 = 1/3, and −b/a = 1/3 ✓. Product = (4/3)(−1) = −4/3, and c/a = −4/3 ✓
✦ Answer: (i) 4, −2 (ii) 1/2, 1/2 (iii) 3/2, −1/3 (iv) 0, −2 (v) √15, −√15 (vi) 4/3, −1 — and all six satisfy sum = −b/a, product = c/a.
Where students slip. In (iii), verifying against a = 6, b = −3, c = −7 because that is the order the question printed. Always rewrite in descending powers of x first — the formulas −b/a and c/a assume standard form.
Another way. Parts (i), (iii) and (vi) also yield to the quadratic formula, and (ii) has discriminant b² − 4ac = 16 − 16 = 0, which is a quick way to spot the repeated zero before factorising.
