Q4) Using Euclid's algorithm, find the HCF of 240 and 228

Euclid's division lemma : Let a and b are two positive integers. There exist unique integers q and r such that

a = bq + r, 0 r < b

Or We can write it as,

Dividend = Divisor × Quotient + Remainder

Work out:

Given integers are 240 and 228. Clearly 240 > 228. Applying Euclid's division lemma to 240 and 228,

⇛ 240 = 228 × 1 + 12

Since, the remainder 12 ≠ 0. So, we apply the division dilemma to the division 228 and remainder 12,

⇛ 228 = 12 × 19 + 0

The remainder at this stage is 0. So, the divider at this stage or the remainder at the previous age i.e 12

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80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol

answer: c

(2, -4)

goes 7 to the

(9, 4)

easy