Using long division method, divide the polynomial
$12 z^{3} + 4 z + 3 z^{2} + 1$ by $4 z + 1$

3 z^{2} + 1

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3 z^{2} - 1

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z^{2} + 1

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None of the above

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Solution

The correct option is A $3 z^{2} + 1$
$3 z^{2} + 1$
$4 z + 1$
$12 z^{3} + 3 z^{2} + 4 z + 1$
$12 z^{3} + 3 z^{2}$

$4 z + 1$
$4 z + 1$
$0$
$(12 z^{3} + 4 z + 3 z^{2} + 1) \div$ $(4 z + 1) = 3 z^{2} + 1$

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Similar questions

\frac{2}{3} z^{4} - \frac{1}{3} z^{2} - 1

\frac{1}{3} z

The degree of the polynomial $3 z^{2} + z + 1$ is 4.

z = 2 - 3 i

z^{2} - 4 z + 13 = 0

4 z^{3} - 3 z^{2} + 169

z_{1} \neq 0

z_{2}

\frac{z_{2}}{z_{1}}

∣ ∣ ∣ \frac{2 z_{1} + 3 z_{2}}{2 z_{1} - 3 z_{2}} ∣ ∣ ∣

x + y + z = x y z

\frac{3 x - x^{3}}{1 - 3 x^{2}} + \frac{3 y - y^{3}}{1 - 3 y^{2}} + \frac{3 z - z^{3}}{1 - 3 z^{2}} = \frac{3 x - x^{3}}{1 - 3 x^{2}} \cdot \frac{3 y - y^{3}}{1 - 3 y^{2}} \cdot \frac{3 z - z^{3}}{1 - 3 z^{2}}

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