Question

if 3x-1 is a factor of polynamial (81x^3-45x^2)+(3a-6) then ais

Answer 1

See the attachment:

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Answer 2

Answer:

The value of a =  $\frac{8}{3}$

Step-by-step explanation:

Given,

3x-1 is a factor of $(81x^3-45x^2)+(3a-6)$

To find,

The value of 'a'

Recall the theorem,

Factor Theorem

If the linear polynomial (x-a) is a factor of p(x), then p(a) = 0

Solution:

Let p(x) be $(81x^3-45x^2)+(3a-6)$

Since 3x -1 is a factor of p(x) we have

3x -1 = 0

x = $\frac{1}{3}$ ⇒ x-$\frac{1}{3}$ = 0

The by factor theorem we have

p($\frac{1}{3}$ ) = 0

$(81(\frac{1}{3}) ^3-45(\frac{1}{3}) ^2)+(3a-6)$ = 0

$(81(\frac{1}{27})-45(\frac{1}{9}) )+(3a-6)$ = 0

3 -5 +3a -6 = 0

3a = 8

a = $\frac{8}{3}$

∴The value of a =  $\frac{8}{3}$

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