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4 years ago

7

X^3(x-3)-(x-3)=(x-3)(?) help

1 answer:

docker41 [41]4 years ago

6 0

Answer:

(x-3)(x^3-1)

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

Step-by-step explanation:

x^3(x-3)-(x-3)=(x-3)(?)

Factor out ( x-3) on the left

(x-3)(x^3-1)=(x-3)(?)

This can be factored further since it is the difference of cubes

a^3 - b^3 = (a−b)(a^2+ab+b^2)

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

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Step-by-step explanation:

<u>A)a1=-2 and d=4</u>

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------------------------------------------

Similarly

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Substituting the given value we get

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-------------------------------------------------

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Substituting the given value we get

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-------------------------------------------------------------------------------------------

<u>B</u><u> $a_n=a_{(n-1)}-8$ with $a_1=10$ </u>

$a_2=a_{(2-1)}-8$

$a_2=a_{1}-8$

Substituting the given value

$a_2= 10-8$

$a_2=2$

---------------------------------------------------------------------

$a_3=a_{(3-1)}-8$

$a_3=a_{2}-8$

Substituting the value

$a_3=2-8$

$a_3= -6$

---------------------------------------------------------------------

$a_4=a_{(4-1)}-8$

$a_4=a_{3}-8$

Substituting the value

$a_4= -6-8$

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-------------------------------------------------------------------------------------------

<u>C) $a_1=3, a_2=1$ </u>

Here the difference is -2

the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{3}=a_{1}+(3-1)d$

$a_{3}=a_{1}+(2)d$

Substituting the value we get

$a_{3}= 3+(2)-2$

$a_{3}= 3-4$

$a_{3}= -1$

------------------------------------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the value we get

$a_{4}= 3+(3)-2$

$a_{4}= 3-6$

$a_{4}= -3$

----------------------------------------------------------------------------------

$a_{5}=a_{1}+(5-1)d$

$a_{5}=a_{1}+(4)d$

Substituting the value we get

$a_{5}= 3+(4)-2$

$a_{5}= 3-8$

$a_{5}= -5$

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