Question

12. What must be added to or subtracted from 15x + 193* + 37x'- 19x2 + 40x – 25 so that it may be exactly divisible by 5r-2x +4?
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Answer 1

Answer

p(x)=g(x)×q(x)+r(x) [Division algorithm for polynomials]

⇒p(x)−r(x)=g(x)×q(x)

It is clear the RHS of the above equation is divisible by g(x). i.e., the divisor.

∴ LHS is also divisible by the divisor.

Therefore, if we subtract remainder r(x)from dividend p(x), then it will be exactly divisible by the divisor g(x)]

Let us divide, 6x4+13x3+13x2+30x+20 by 3x2+2x+5

we get, quotient q(x)=2x2+3x−1remainder r(x)=17x+25

∴ If we subtract 17x+25 from 6x4+13x3+13x2+30x+20, it will be exactly divisible by (3x2+2x+5).

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Answer 2

Answer:

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Step-by-step explanation:

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