Question

Can someone help me understand this problem better? please?..

Match the pairs of values of f(x) and g(x) with the corresponding values of h(x) if h(x) = f(x) ÷ g(x).

Answer 1

Pairs: Tiles:

h(x) = x + 5 ? f(x) = x2 − 9, and g(x) = x − 3

h(x) = x + 3 ? f(x) = x2 − 4x + 3, and g(x) = x − 3

h(x) = x + 4 ? f(x) = x2 + 4x − 5, and g(x) = x − 1

h(x) = x − 1 ? f(x) = x2 − 16, and g(x) = x − 4

Answer 2

Let's look at the first pair of "tiles"\[\large\rm h(x)\quad=\frac{f(x)}{g(x)}\quad=\frac{x^2-9}{x-3}\]Looks something like that, ya?
The top can factor using our Difference of squares Formula,\[\large\rm a^2-b^2=(a-b)(a+b)\]

Answer 3

correct @zepdrix

Answer 4

Do you see how the top will factor? :)
Rewrite 9 as 3^2.\[\large\rm x^2-9\quad=x^2-3^2\]

Answer 5

\[\large\rm x^2-3^2\quad=?\]Any ideas? ^^

Answer 6

no thats why i asked for help lol @zepdrix

Answer 7

\[\large\rm a^2-b^2=(a-b)(a+b)\]\[\large\rm x^2-3^2\quad=?\]See how the x is a in the formula, 3 is the b...

Answer 8

\[\large\rm h(x)\quad=\frac{f(x)}{g(x)}\quad=\frac{\color{orangered}{x^2-9}}{x-3}\]So applying our formula to this numerator,\[\large\rm h(x)\quad=\frac{f(x)}{g(x)}\quad=\frac{\color{orangered}{(x-3)(x+3)}}{x-3}\]And then you can cancel out a common factor in the numerator/denominator.