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AnilSinghBora
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Factorize:(ii) x3–3×2–9x–5 Q.5(2)

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I don’t know how to solve this problem in simple way of Polynomials chapter of exercise 2.4 of class 9th math. Give me the best answer of this question  Factorize:(ii) x3–3×2–9x–5

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  1. Let p(x) = x3–3x2–9x–5

    Factors of 5 are ±1 and ±5

    By trial method, we find that

    p(5) = 0

    So, (x-5) is factor of p(x)

    Now,

    p(x) = x3–3x2–9x–5

    p(5) = (5)3–3(5)2–9(5)–5

    = 125−75−45−5

    = 0

    Therefore, (x-5) is the factor of  p(x)

    Ncert solutions class 9 chapter 2-2

    Now, Dividend = Divisor × Quotient + Remainder

    (x−5)(x2+2x+1) = (x−5)(x2+x+x+1)

    = (x−5)(x(x+1)+1(x+1))

    = (x−5)(x+1)(x+1)

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