If n > 0 and 9x^2 + kx + 36 = (3x + n)^2 for all values of x, what’s the value of k-n?
Expand the expression (3x+n)^2 and because these expressions are equal to each other for all values of x, the coefficients of the quadratic terms must be equal. I will leave the rest for you to finish.
(3x+n)^2=9x^2+6xn+n^2
9x^2 + kx + 36=9x^2+6xn+n^2
kx=6xn ==> k=6n
36=n^2 ===> 6=n
k =6(6) = 36
k-n = 30
36-6 = 30
I’m just going to repeat the last answer, but with words.
The point is to do this quickly and correctly. Break down the original quadratic to two equal factors.
9x^2 breaks down to (3x )(3x )
36 breaks down to 6 and 6. Since all the signs are positive, they stay positive in the factors.
So you have (3x+6) ^2
k = 2(3)(6) or 36. n=6. (FOIL. 3(6) for the inners, again for the outers, sum = 36)
k-n=30
Or, if you are feeling sneaky:
Since it is true for all x, try x = 0. You get 36 = n^2 so n = 6 [they said n was positive]
Then try x=1. You get 9 + k + 36 = 81. So k = 36.
This kind of approach works on all of those problems where they give you two polynomials that are equal for all x but one is factored out and the other not and they ask you about the coefficients.