Polynomials Test - 2

Using remainder theorem:
f(x) = x³ + 7x² + 14x + 14

When f(x) is divided by (x + 4), the remainder obtained will be f(-4).

f(-4) = (-4)³ + 7(-4)² + 14(-4) + 14
= -64 + 112 - 56 + 14
= 6

f(x) = px⁴ + qx³ + rx² + sx + t

x² - 25 = (x – 5) (x + 5)

f(-5) = f(5) = 0

f(-5) = p(-5)⁴ + q(-5)³ + r(-5)² + s(-5) + t

= 625p - 125q + 25r - 5s + t (Equation I)

f(5) = p(5)⁴ + q(5)³ + r(5)² + s(5) + t

= 625p + 125q + 25r + 5s + t (Equation II)

Equating equations (I) and (II), we get:

625p - 125q + 25r - 5s + t = 625p + 125q + 25r + 5s + t

250q + 10s = 0

25q + s = 0

Total biscuits in a pack = (4b + 3a)
Total biscuits in a box = (4b - 3a)(4b + 3a) = 16b² - 9a² (Using identity: (x + y) (x - y) = x² - y²)
Total biscuits in the store = (16b² + 9a²) (16b² - 9a²) = 256b⁴ - 81a⁴ (Using identity: (x + y) (x - y) = x² - y²)

f(x) = x³ - 2x² - 12x + 16
g(x) = x – 4

The remainder obtained will give us the number of candies left over.
Using remainder theorem:
Remainder = f(4)

f(4) = (4)³ - 2 (4)² - 12 (4) + 16
= 64 - 32 - 48 + 16
= 0

f(x) = x³ + x² - 2ax - 2b
In case no sweet is left over, the remainder is 0.
f(2) = 2³ + 2² - 4a - 2b = 0
2a + b = 6 ...(i)

Also,
f(-1) = (-1)³ + (-1)² + 2a - 2b = 0
a = b ...(ii)

Solving (i) and (ii), we get:
a = b = 2

If x = 5, then
f(5) = 5³ + 5² - 2(2)(5) - 2(2)
= 125 + 25 – 20 – 4 = 126

She has 126 sweets.