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
As pointed out, the factorized form of the equation
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 -6
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)
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.
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,