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2 years ago

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

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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3 years ago

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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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 $(81, 4 )$

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3 years ago