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Svetach [21]
3 years ago
10

I’m doing edg unit test what is the answer to. Which shows one way to determine the factors of 4x3 + x2 – 8x – 2 by grouping?

Mathematics
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

(4x + 1)(x² - 2)

Step-by-step explanation:

Given

4x³ + x² - 8x - 2 ( factor the first/second and third/fourth terms )

= x²(4x + 1) - 2(4x + 1) ← factor out (4x + 1) from each term

= (4x + 1)(x² - 2)

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3+7(0.7+1.3)
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5 0
2 years ago
If x = a sin α, cos β, y = b sin α.sin β and z = c cos α then (x²/a²) + (y²/b²) + (z²/c²) = ?​
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\large\underline{\sf{Solution-}}

<u>Given:</u>

\rm \longmapsto x = a \sin \alpha  \cos \beta

\rm \longmapsto y = b \sin \alpha  \sin \beta

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Therefore:

\rm \longmapsto \dfrac{x}{a}  = \sin \alpha  \cos \beta

\rm \longmapsto \dfrac{y}{b}  = \sin \alpha  \sin \beta

\rm \longmapsto \dfrac{z}{c} = \cos \alpha

Now:

\rm =  \dfrac{ {x}^{2} }{ {a}^{2}} +  \dfrac{ {y}^{2} }{ {b}^{2} } +  \dfrac{ {z}^{2} }{ {c}^{2} }

\rm =  { \sin}^{2} \alpha  \cos^{2}  \beta   +  { \sin}^{2} \alpha  \sin^{2} \beta  +  { \cos}^{2} \alpha

\rm =  { \sin}^{2} \alpha  (\cos^{2}  \beta   +  \sin^{2} \beta  )+  { \cos}^{2} \alpha

\rm =  { \sin}^{2} \alpha \cdot1+  { \cos}^{2} \alpha

\rm =  { \sin}^{2} \alpha + { \cos}^{2} \alpha

\rm = 1

<u>Therefore:</u>

\rm \longmapsto\dfrac{ {x}^{2} }{ {a}^{2}} +  \dfrac{ {y}^{2} }{ {b}^{2} } +  \dfrac{ {z}^{2} }{ {c}^{2} }  = 1

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Answer:

to expand, use the distributive property. therefore, we get:

6x - 24y

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Answer:

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Step-by-step explanation:

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Answer:

Line M - neither

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