# Find f+g, f-g, f*g, and f divided by g.  f(x)=6x-9  g(x)=3x+2PLEASE SHOW ALL WORK!!!!

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that addition of two polynomial functions follows several rules, hence you need to collect like terms meaning you need to add the terms containing x and the constant terms separately such that:

`f(x) + g(x) = (6x - 9) + (3x + 2)`

Collecting the like terms yields:

`f(x) + g(x) = (6x + 3x) + (-9 + 2)`  (the constant terms keep the signs they have)

`f(x) + g(x) = 9x - 7`  (adding a negative number to a positive number means subtracting the number that is smaller in absolute value from the number that is larger in absolute value).

You may perform the addition of two polynomials using the vertical format such that:

`6x - 9 +`

`3x + 2`

----------

`9x - 7`

Notice that either you select the horizontal format or the vertical format, you will get the same binomial 9x - 7.

You need to remember that subtraction of two polynomial function works the same as the addition hence:

`f(x)- g(x) = (6x - 9)- (3x + 2)`

You need to open the brackets considering the minus sign such that:

`f(x)- g(x) = 6x - 9 - 3x - 2`

`f(x)- g(x) = (6x - 3x) - 9 - 2`

`` `f(x)- g(x) = 3x - 11`

You need to multiply two terms polynomial hence, using the horizontal format, you need to distribute the binomial (6x - 9) to each term in (3x + 2) such that:

`f(x)*g(x) = (6x - 9)*(3x + 2)=gtf(x)*g(x) = (6x - 9)*(3x) + (6x - 9)*(2)`

You need to distribute 3x to each term in 6x - 9, then you need to distribute 2 to each term in 6x - 9 such that:

`f(x)*g(x) = 3x*6x + 3x*(-9) + 2*6x + 2*(-9)`

`f(x)*g(x) = 18x^2 - 27x + 12x - 18`

Collecting like terms yields:

`f(x)*g(x) = 18x^2 - 15x - 18`

You need to perform the division of two terms polynomial such that:

`(f(x))/(g(x)) = (6x - 9)/(3x + 2)`

You need to split the division in two fractions (there are only two terms to numerator) such that:

`(f(x))/(g(x)) = (6x)/(3x + 2) +(- 9)/(3x + 2)`

`` `(6x) = 2*3x`

Considering the fraction `(3x)/(3x + 2)`  you may notice that there is missing a constant term to numerator to become the same binomial as denominator, hence you will add the constant term 2 to numerator but you will also subtract the term 2 to preserve the fraction such that:

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

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

Hence,you may write the fraction `(6x)/(3x + 2)`  such that `2*(1 - 2/(3x + 2))` .

Substituting  `2*(1 - 2/(3x + 2))`  for `(6x)/(3x + 2)`  in division yields:

`(f(x))/(g(x)) = 2 - 4/(3x + 2) - 9/(3x + 2)`

`` Collecting like terms yields:

`(f(x))/(g(x)) = 2 - 13/(3x + 2)`

`` `(6x - 9)/(3x + 2) = 2 - 13/(3x + 2)`

Hence, dividing f(x) by g(x) yields the quotient `q(x) = 2`  and the reminder `r(x) = -13` .

Hence, evaluating addition of f(x) and g(x) yields `f(x) + g(x) = 9x - 7` , subtraction`f(x)- g(x) = 3x - 11` , multiplication `f(x)*g(x) = 18x^2 - 15x - 18 `  and division `(f(x))/(g(x)) = 2 - 13/(3x + 2).`