Question

If 3, -2 and 1 are the zeroes of Cubic polynomial find the polynomial.​

Answer 1

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Answer 2

Answer:

∝ = 3, β = -2, r = 1

= ∝ + β + r = (3) + (-2) + (1)

= 3 -2 + 1

= 4 - 2

= 2

= ∝β + βr + ∝r = (3) (-2) + (1) + (3) (1)

= -6 + (-2) + 3

= -8 +3

= -5

= ∝βr = (3) (-2) (1)

= -6

= k [$x$³ - (∝ + β + r) $x^2$ + (∝β + βr + ∝r) $x$ - ∝β]

=  k [$x$³ - (2) $x^2$ + (-5)$x$ - (-6)

=  k [$x$³ - 2)$x^2$ - 5$x$ + 6

= k = 1, p ($x$) = $x^3$ + 2$x^2$ - 5$x$ + 6

= $x$ = 1 => f(1) = (1)³ - 2$x$(1)² - 5(1) + 6

= 1 - 2 - 5 + 6

= +4 -4

= 0

∴ (x -2) is a factor of p ($x$) = $x^3$ + 2$x^2$ - 5$x$ + 6

  $x$ - 2 ) $x^3$ + 2$x^2$ - 5$x$ + 6 ( $x$ + 4$x$ + 3

            $x$³ + 2$x$²

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                 +4$x$² - 5$x$

                  -4$x$² - 8$x$

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                              3$x$ + 6

                              3$x$ + 6

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                                   0

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= ($x$²+ 4x + 3)

= $x$² - $x$ - 3$x$ + 3

= $x$ ($x$ - 1) -3 ($x$ - 1)

= ($x$ - 3) ($x$ -3) = 0

= ($x$ - 3) = 0

= $x$ = 3

= ($x$ -1) = 0

= $x$ = 1