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Andreyy89
3 years ago
9

Which of the following are factors of x^4 – 16? Select all that apply.

Mathematics
2 answers:
Alex73 [517]3 years ago
7 0

Answer:

A and D.

Step-by-step explanation:

x^4 -16

(x^3+2x^2+4x+8)(x-2)

(x^2+4)(x+2)(x-2)

nekit [7.7K]3 years ago
5 0

<h2><em>Answer:</em></h2><h2><em>x^</em><em>2</em><em>+</em><em>4</em><em> </em><em>and </em><em>X+</em><em>2</em></h2>

<em>Solution,</em>

<em>x^</em><em>4</em><em>-</em><em>1</em><em>6</em>

<em>=</em><em> </em><em>(</em><em>X^</em><em>2</em><em>)</em><em>^</em><em>2</em><em>-</em><em>(</em><em>4</em><em>)</em><em>^</em><em>2</em>

<em>Use </em><em>the </em><em>formula(</em><em>a^</em><em>2</em><em>-</em><em>b</em><em>^</em><em>2</em><em>)</em>

<em>=</em><em> </em><em>(</em><em>x^</em><em>2</em><em>+</em><em>4</em><em>)</em><em>(</em><em>x^</em><em>2</em><em>-</em><em>4</em><em>)</em>

<em>=</em><em>(</em><em>x^</em><em>2</em><em>+</em><em>4</em><em>)</em><em>(</em><em>x^</em><em>2</em><em>-</em><em>2</em><em>^</em><em>2</em><em>)</em>

<em>=</em><em>(</em><em>x^</em><em>2</em><em>+</em><em>4</em><em>)</em><em>(</em><em>X+</em><em>2</em><em>)</em><em>(</em><em>x-2)</em>

<em>Hope </em><em>it</em><em> helps</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

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Use lagrange multipliers to find the point on the plane x â 2y + 3z = 6 that is closest to the point (0, 2, 4).
Arisa [49]
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but since \sqrt{f(x,y,z)} and f(x,y,z) share critical points, we can instead consider the problem of optimizing f(x,y,z) subject to x-2y+3z=6.

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with partial derivatives (set equal to 0)

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L_y=2(y-2)-2\lambda=0\implies y=2+\lambda
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which gives the critical point

x=-\dfrac17,y=\dfrac{16}7,z=\dfrac{25}7

We can confirm that this is a minimum by checking the Hessian matrix of f(x,y,z):

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

\mathbf H is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.

At this point, we get a distance from (0, 2, 4) of

\sqrt{f\left(-\dfrac17,\dfrac{16}7,\dfrac{25}7\right)}=\sqrt{\dfrac27}
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