Question

factorise 1/3 c^3 - 2c - 9

Answer 1

Answer:

answer is $c = 9, -3$

Step-by-step explanation:

Given :- Equation to be factorize :- $\frac{1}{3}c^2 - 2c -9$

To find :- Factors of equation .

Solution :-

Step 1) write the equation ,    $\frac{1}{3} c^2 - 2c -9$

Step 2) suppose the equation is equal to 0 , then only we will be able to find the factors of eqn .

$\frac{1}{3}c^2 - 2c -9 =0$

Step 3) multiply whole equation by 3 ,

we get , $c^2 - 6c-27 = 0$

Step 4) Now factorize the equation as ,

$c^2 - 9c+ 3c -27 = 0 \\c(c-9) + 3 (c-9) = 0 \\(c-9) * ( c+3 ) = 0 \\c = 9 , -3$

Therefore , answer is $c = 9, -3$

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Answer 2

Answer:

9 and -3 are the value of "c" .

Step-by-step explanation:

Explanation:

Given, $\frac{1}{3} c^2 - 2c -9$

Factorization- The factorization method allows us to simplify any algebraic or quadratic equation by representing the equations as the product of factors rather than by expanding the brackets.

Step 1:

From the question we have,

$\frac{1}{3} c^2 - 2c -9$ = 0

And this can be written as $\frac{1}{3} c^2 - \frac{2c}{1} -\frac{9}{1} = 0$.

LCM of 3 and 1 is 3.

So, $\frac{c^2 - 6c - 27 }{3} = 0$ ⇒$c^2 - 6c - 27 = 0$

Now by factorization method,

⇒$c^2 - 9c +3c - 27$ = 0

⇒c(c - 9) + 3(c - 9) = 0

⇒(c - 9 )(c + 3) = 0

Now, c - 9 = 0  and c + 3 = 0

c = 9 and c = -3

So, the value of 'c' = 9 and -3

Final answer:

Hence, after factorization value of c are 9 and -3 .

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