# The function f(x) = x^3+ax+6 is such that when divided by (x-3) the remainder is 12. Factorize f(x) by showing the value of a = 7

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Polynomial remainder theorem states that if f(x) is divided by (x-a), the remainder is given by f(a).

Here, `f(x) =` `x^3+ax+6` is divided by (x-3) and the remainder is 12.

Thus, `f(3)=12`

`rArr 3^3+3a+6=12`

`rArr 3a=-21`

`rArr a=-7`

Hence, `f(x) = x^3-7x+6` which is to be factorized.

By trial and error method we find that `f(-3)=0` . So, `(x+3)` is a factor of the given function. Thus,

`x^3-7x+6 `

`=x^3+3x^2-3x^2-9x+2x+6`

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

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

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

`=(x+3){x(x-2)-1(x-2)}`

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

**Sources:**

As pointed out, the factorized form of the equation

`f(x)=x^3 +ax+6`

requires that a= -7 and not 7.

This can also be established using synthetic division because we know we have a remainder of 12 when using (x-3) and thus x=3:

Stat with the x=3 and then multiply using the co-efficients (ie the numbers in front of the x-es and the constant which is 6) of `x^3 +ax + 6` which are 1;0;a and 6. Take care to include the zero (0) because there is no `x^2` which means that the co-efficient of `x^2=0` and there is a 1 in front of `x^3` .

When doing synthetic division, you will complete line 3 and carry that answer forward to the next number's line 2, such that after bringing down the 1, `3 times1 = 3` which appears in line 2 under the zero. Then add the 0+3 which gives you the 3 in the 3rd line. Now multiply that 3 by the first 3 = 9 which is carried to the 2nd line under the a. Now add the 9 to the a. Then multiply the 9+a by the first 3 to continue the pattern and then add the 6 to the 3(9+a). In other words to create a new number multiply the 3 (in this case) by the answer you get in the third line each time.

3 / 1....0.....a.........6

..... 3 9 3(9+a)

...1 ....3...(9+a)...6+ 3(9+a)

We already know that the remainder is 12 and our last commuted result from our synthetic division is the remainder.

`therefore 6+3(9+a) = 12`

`therefore 27 + 3a = 12 - 6`

`therefore 3a = 6-27`

`therefore a = -21/3`

`therefore a= - 7`

Now inserting the value of **a=-7** into `x^3+ax+6` we get

`f(x) = x^3 -7x+6` If we make x=1

`therefore f(x) = 1^3 -7(1) +6` which =0 and we therefore know that 1 is a root or x-value which means that (x-1) is a factor. Use synthetic division again, remembering to insert the zero as we have no x^2 and using the 1 as the factor and the -7 instead of a. Bring down the 1 :

1/...1....0....-7....6

..... 1 1 -6

1.....1....-6....0

This shows there is no remainder (0) and gives us the co-efficients for our remaining equation so we have (x-1) and now we also have `(1x^2 +1x -6)` . Note where the co-efficients came from above (1...1...-6).

We know from the previous solution that this will give us (x-2)(x+3) so factorized,

**having proven that a= -7 ****we get (x-1)(x-2)(x+3) **