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Olenka [21]
2 years ago
12

Factor the following expression completely. 16x^5-x^3 A. B. C. D.

Mathematics
2 answers:
laiz [17]2 years ago
8 0

Answer:

The factor of the provided expression are: x^3(4x-1)(4x+1)

Step-by-step explanation:

Consider the provided expression.

16x^5-x^3

Here the Greatest common factor in the above expression is x³.

The above expression can be written as:

x^3(16x^2-1)

x^3((4x)^2-1^2)

Now use the difference of the square property: a^2-b^2=(a+b)(a-b)

By using the above property we can rewrite the provided expression as shown:

x^3(4x-1)(4x+1)

Hence, the factor of the provided expression are: x^3(4x-1)(4x+1)

Ket [755]2 years ago
6 0

Answer:

x^3(2x+1)(2x-1)

Step-by-step explanation:

first take common then use formula of a^2-b^2

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\bigstar\:{\underline{\sf{In\:right\:angled\:triangle\:ABC\::}}}\\\\

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⠀⠀⠀

\bf{\dag}\:{\underline{\frak{By\:using\:Pythagoras\: Theorem,}}}\\\\

\star\:{\underline{\boxed{\frak{\purple{(Hypotenus)^2 = (Perpendicular)^2 + (Base)^2}}}}}\\\\\\ :\implies\sf (AB)^2 = (AC)^2 + (BC)^2\\\\\\ :\implies\sf (AB)^2 = (AB)^2 = (7)^2 = (4)^2\\\\\\ :\implies\sf (AB)^2 = 49 + 16\\\\\\ :\implies\sf (AB)^2 = 65\\\\\\ :\implies{\underline{\boxed{\pmb{\frak{AB = \sqrt{65}}}}}}\:\bigstar\\\\

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

☆ Now Let's find value of sin A, cos A and tan A,

⠀⠀⠀

  • sin A = Perpendicular/Hypotenus = \sf \dfrac{4}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{4 \sqrt{65}}{65}}

⠀⠀⠀

  • cos A = Base/Hypotenus = \sf \dfrac{7}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{7 \sqrt{65}}{65}}

⠀⠀⠀

  • tan A = Perpendicular/Base = {\sf{\pink{\dfrac{4}{7}}}}

⠀⠀⠀

\therefore\:{\underline{\sf{Hence,\: {\pmb{Option\:A)}}\:{\sf{is\:correct}}.}}}

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