The minimum value of the polynomial.
$p (x) = 3 x^{2} - 5 x + 2$

- \frac{1}{6}

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\frac{1}{6}

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\frac{1}{12}

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- \frac{1}{12}

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Solution

The correct option is D $- \frac{1}{12}$
$p (x) = 3 x^{2} - 5 x + 2$
$= 3 (x 2 - \frac{5}{3} x + \frac{2}{3})$
$= 3 [x^{2} - \frac{5}{3} x + {(\frac{5}{6})}^{2} - {(\frac{5}{6})}^{2} + \frac{2}{3}]$
$= 3 [{(x - \frac{5}{6})}^{2} - \frac{25}{36} + \frac{2}{3}]$
$= 3 [{(x - \frac{5}{6})}^{2} + \frac{- 25 + 24}{36}]$
$= 3 [{(x - \frac{5}{6})}^{2} - \frac{1}{36}] = 3 {(x - \frac{5}{6})}^{2} - \frac{1}{12}$

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Similar questions

p (x) = 3 x^{2} - 5 x + 2

Statement1: The polynomial $P (x) = 4 x^{3} - 3 x^{2} + 5 x - 6$ when divided by $x - 1$ gives zero as the remainder.

Statement2: $(x - 1)$ is a factor of the polynomial $P (x) = 4 x^{3} - 3 x^{2} + 5 x - 6$ .

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i)

(ii)

(iii)

P (x + 2) - P (x) = x^{3} + 3 x^{2} + x + 1

\frac{1}{2}, 3

\frac{1}{2}, - 3

\frac{- 1}{2}, 3

1, \frac{- 1}{2}

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