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Laws of Exponents for Real Numbers

Simplify: √...

Question

Simplify: $\sqrt{4 x^{4} + 8 x^{3} + 16 x^{2} + 12 x + 9}$ (Use division method).

A

∣ ∣ 2 x^{2} + 2 x + 3 ∣ ∣

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B

2 x^{2} + 2 x + 3

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C

∣ ∣ 2 x^{2} - 2 x + 3 ∣ ∣

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D

2 x^{2} - 2 x - 3

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Solution

The correct option is A $∣ ∣ 2 x^{2} + 2 x + 3 ∣ ∣$
$\div$ $2 x^{2} + 2 x + 3$ <= Quotient
$2 x^{2}$
$4 x^{4} + 8 x^{3} + 16 x^{2} + 12 x + 9$
$4 x^{4}$
$4 x^{2} + 2 x$ $8 x^{3} + 16 x^{2}$
$8 x^{3} + 4 x^{2}$
$4 x^{2} + 4 x + 3$ $12 x^{2} + 12 x + 9$
$12 x^{2} + 12 x + 9$
Remainder $\Rightarrow 0$
The square root of $4 x^{4} + 8 x^{3} + 16 x^{2} + 12 x + 9$ is $∣ ∣ 2 x^{2} + 2 x + 3 ∣ ∣$

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Similar questions

Q. Finding the square root of the polynomial

4 x^{4} + 8 x^{3} + 8 x^{2} + 4 x + 1.

[4 x^{4} + 8 x^{3} + 8 x^{2} + 4 x + 1]

can be simplified to

(2 x^{2} + 2 x + 1)^{2}

(\frac{x^{2}}{3} - \frac{4 x}{7} + 11) - (\frac{x}{7} - 3 + 2 x^{2}) - (\frac{2 x}{7} - \frac{2 x^{2}}{3} + 2)

Q. Solve the equations:

\frac{3 x^{4} + x^{2} - 2 x - 3}{3 x^{4} - x^{2} + 2 x + 3} = \frac{5 x^{4} + 2 x^{2} - 7 x + 3}{5 x^{4} - 2 x^{2} + 7 x - 3}

\sqrt{2 x^{2} - 2 x + 1} = 2 x - 3

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