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Setler79 [48]
2 years ago
5

GIVING BRAINLIEST!!!Show that(Factor it) 4x^4-12x^3+22x^2-24x+26=(x-1)^2[(2x-1)^2+9]​

Mathematics
1 answer:
-BARSIC- [3]2 years ago
3 0

Answer:

  • See below

Step-by-step explanation:

<u>Given equation:</u>

  • 4x⁴ - 12x³ + 22x² - 24x + 26 = (x-1)²[(2x-1)² + 9]​

<u>Expand the right side and compare with left side:</u>

  • (x - 1)²[(2x - 1)² + 9]​ =
  • (x² - 2x + 1)(4x² - 4x + 1 + 9) =
  • (x² - 2x + 1)(4x² - 4x + 10) =
  • x²(4x² - 4x + 10) - 2x(4x² - 4x + 10)  + (4x² - 4x + 10)  =
  • 4x⁴ - 4x³ + 10x² - 8x³ + 8x² - 20x  + 4x² - 4x + 10 =
  • 4x⁴ - 12x³ + 22x² - 24x + 10

As we see LHS is different from RHS and the difference is 26 ≠ 10, all the other terms are equal

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Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)
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Answer:

Considering the given equation y = log_{3}x\\

And the ordered pairs in the format (x, y)

I don't know if it is log of base 3 or 10, but I will assume it is 3.

For (\frac{1}{3}, a_{0} )

x=\frac{1}{3}

y=a_{0}

y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1

So the ordered pair will be (\frac{1}{3}, -1 )

For (1, a_{1} )

x=1

y=a_{1}

y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0

So the ordered pair will be (1, 0 )

For (3, a_{2} )

x=3

y=a_{2}

y = log_{3}x\\y = log_{3}3\\y = 1

So the ordered pair will be (3, 1 )

For (9, a_{3} )

x=9

y=a_{3}

y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2

So the ordered pair will be (9, 2 )

For (27, a_{4} )

x=27

y=a_{4}

y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3

So the ordered pair will be (27, 3 )

For (81, a_{5} )

x=81

y=a_{5}

y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4

So the ordered pair will be (81, 4 )

4 0
3 years ago
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