0
$\begingroup$

My studies into graphs and models following examples from Khan Academy has helped me on my goal to learn how to chart and model via the quadratic formula.

However, while I have been successful in understanding addition, subtraction and multiplication of 'g' and 'f':

Say

$f(x) = 2x^2 + 15x -8$

$g(x) = x^2 + 10x + 16$

I have been unsuccessful in understanding division,

Where $f(x) / g(x)$

Would one start by grouping the functions into their own families?, for example:

$2x^2 / x^2$

$15x / 10x$

$-8 / 16$

I only came to this conclusion as with multiplication you times as so : $(A+B) \cdot (C+D) = AC + AD + BC + BD$

However from the example on Khan Ac. this seems wrong, it appears that division departs from any systematic way to find the answer, and that Khans division explanation is not as clear as his others on +, -, and * (appears to me!)

I decided to try and explore division once more in its most simple form to see if I was missing something.

$\frac23= 0.666666666...$

$\frac24 = 0.5$

Here clearly, certain division of numbers produces what seems like a simple solution into irrational numbers, If I were to divide a cake the same way I would think of each piece as a ratio, however in decimals these numbers are not clearly defined.

And when one starts dividing multiple factors as with above, how does one find the correct way to divide a quadratic equation, without relying on cheap tricks.

Any easily accessible writing into the theory of division would also be helpful, thank you!

$\endgroup$
2
$\begingroup$

Both factor, $$ x^2 + 10 x + 16 = (x + 2)(x + 8), $$ $$ 2x^2 + 15 x - 8 = (2x-1)(x+8). $$ So your fraction is undefined at $x = -8,$ otherwise it is the same as $$ \frac{x + 2}{2x-1}. $$ The graph of this has both horizontal and vertical asymptotes; it is a hyperbola.

The reason I knew they factored was that the discriminants came out squares, $$ 10^2 - 4 \cdot 1 \cdot 16 = 36 = 6^2, $$ $$ 15^2 - 4 \cdot 2 \cdot (-8) = 289 = 17^2. $$

It would also help you to know how to find the greatest common factor of two polynomials; it is $(x+8)$ in this case. One caution, it may be necessary to allow rational coefficients to find a gcd. Not sure what else to say.

Well, why not. To put an ordinary fraction into lowest terms, you find the greatest common divisor of numerator and denominator, then divide that out of both top and bottom. that is what they are trying to show you here. Let us use a symbol $$ g = \gcd(2x^2 + 15 x - 8, \; \; x^2 + 10 x + 16) $$ First we find $$ 2x^2 + 15 x - 8 - 2 \cdot ( x^2 + 10 x + 16) = -5x-40 = -5 (x+8) $$ So, possibly with rational quotient, $$ g = \gcd(x^2 + 10 x + 16, \; \; x+8). $$ But then $$ x^2 + 10 x + 16 - (x+2)(x+8) = 0, $$ so $$ g = x+8. $$

$\endgroup$
1
$\begingroup$

To find the quotient $$\frac{f(x)}{g(x)} = \frac{2x^2 + 15x - 8}{x^2 + 10x + 16}$$ we must find what we need to multiply $x^2$ by to obtain $2x^2$. Clearly, this is $2$. $$2x^2 + 15x - 8 = 2(x^2 + 10x + 16) - 5x - 40$$ Since $-5x - 40$ has smaller degree than $x^2 + 10x + 16$, $-5x - 40$ is the remainder. Hence, the quotient is $$\frac{2x^2 + 15x - 8}{x^2 + 10x + 6} = 2 + \frac{-5x - 40}{x^2 + 10x + 16}$$

Check: We multiply the quotient by the divisor to show that it equals the dividend. \begin{align*} (x^2 + 10x + 16)\left(2 + \frac{-5x - 40}{x^2 + 10x + 16}\right) & = 2(x^2 + 10x + 16) - 5x - 40\\ & = 2x^2 + 20x + 32 - 5x - 40\\ & = 2x^2 + 15x - 8 \end{align*}

Note: Since \begin{align*} \frac{f(x)}{g(x)} & = \frac{2x^2 + 15x - 8}{x^2 + 10x + 16}\\ & = \frac{2x^2 + 16x - x - 8}{x^2 + 2x + 8x + 16}\\ & = \frac{2x(x + 8) - 1(x + 8)}{x(x + 2) + 8(x + 2)}\\ & = \frac{(2x - 1)(x + 8)}{(x + 8)(x + 2)}\\ & = \frac{2x - 1}{x + 2} \end{align*} provided that $x \neq -8$, we could instead perform the simpler division $$\frac{2x - 1}{x + 2}$$ in this case. $$2x - 1 = 2(x + 2) - 5$$ Observe that if we multiply each side of the equation $2x - 1 = 2(x + 2) - 5$ by $x + 8$, we obtain $2x^2 + 15x - 8 = 2(x^2 + 10x + 16) - 5x - 40$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.