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in Factorization by (31.2k points)
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Factories:

i. 14x3 + 21x 4y – 28x2y2

ii. - 5 – 10t + 20t2

1 Answer

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Best answer

(i) Let’s find out the HCF of 14x3, 21x4y and 28x2y2

7x2 is the highest common factor of 14x3, 21x4y,28x2y2

So,

14x3 + 21x4y – 28x2y2 = 7x2(2x + 3x2y – 4y2)

5 is the highest common factor of 5, 10t and 20t2.

So,

- 5 – 10t + 20t2 = - 5(1 + 2t – 4t2)

(Note: As we have learned in the previous chapter when we multiplied – sign with – sign it become +)

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