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X3 − 6x2 + 3x + 10 - CBSE Class 9 - Mathematics

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Question

x3 − 6x2 + 3x + 10

Solution 1

Let f(x) = x3 − 6x2 + 3x + 10 be the given polynomial.

Now, putting  x = -1we get

`f(-1) = (-1)^3 - 6(1)^2 + 3(-1) + 10`

             ` = -1 -6 -3 + 10`

            ` = -10 + 10`

            ` = 0`

Therefore, (x+1)is a factor of polynomial f(x).

Now,

\[f(x) = x^3 - 7 x^2 + x^2 + 10x - 7x + 10\]

`f(x) = x^2 (x + 1) - 7x(x+1)+10(x +1)`

        ` =(x +1){x^2 - 7x + 10} `

        ` = (x+1){x^2 - 5x - 2x + 10}`

        ` = (x+1)(x-5)(x-2)`

Hence, (x+1),(x-2) and (x-5) are the factors of the polynomial f(x).

Solution 2

Let f(x) = x3 − 6x2 + 3x + 10 be the given polynomial.

Now, putting  x = -1we get

`f(-1) = (-1)^3 - 6(1)^2 + 3(-1) + 10`

             ` = -1 -6 -3 + 10`

             ` = -10 + 10`

             ` = 0`

Therefore, (x+1)is a factor of polynomial f(x).

Now,

\[f(x) = x^3 - 7 x^2 + x^2 + 10x - 7x + 10\]

`f(x) = x^2 (x + 1) - 7x(x+1)+10(x +1)`

        ` =(x +1){x^2 - 7x + 10} `

        ` = (x+1){x^2 - 5x - 2x + 10}`

        ` = (x+1)(x-5)(x-2)`

Hence,(x+1),(x-2) and (x-5) are the factors of the polynomial f(x).

  Is there an error in this question or solution?

APPEARS IN

 RD Sharma Solution for Mathematics for Class 9 by R D Sharma (2018-19 Session) (2018 to Current)
Chapter 6: Factorisation of Polynomials
Ex.6.50 | Q: 3 | Page no. 32

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Solution X3 − 6x2 + 3x + 10 Concept: Factorisation of Polynomials.
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