Hi I am trying to simplify the following I found online

$9\left(\dfrac{x^2-15x+50}{84}\right) + -12\left(\dfrac{x^2-8x-20}{-35}\right) + 33\left(\dfrac{x^2-3x-10}{60}\right)$


$= x^2 -6x -7$

Working through this I thought I could multiply the top by the divide for each part of the polynomial. For example

$x^2 \cdot 84 = 84x^2$

$-15x \cdot 84 = 1260x$

$+50 \cdot 84 = 4200$

Then to remove the outside multiply times again.

$84x^2 \cdot 9 = 756x^2$

$-1260x \cdot 9 = 11340x$

$4200 \cdot 9 = 37800$

Do this for all of the parts and then simplify down to the expected

$= x^2-6x-7$

It did not seem to work though unless I was missing something or made a mistake.

Is this the expected method of doing this?

I came up with $3156x^2 - 20640x - 27420$

But I cannot see how that would simplify to the expected.

  • 1
    $\begingroup$ Welcome to Mathematics Stack Exchange. You don't want to multiply by $84$; you want to divide by $84$. Try finding a common denominator $\endgroup$ – J. W. Tanner May 1 at 17:00
  • $\begingroup$ You have $Ax^2+Bx+C=ax^2+bx+c\iff A=a,B=b,C=c$ so you can verify separately three arithmetic sums. $\endgroup$ – Piquito May 1 at 17:42

To simplify $9\left(\dfrac{x^2-15x+50}{84}\right) + -12\left(\dfrac{x^2-8x-20}{-35}\right) + 33\left(\dfrac{x^2-3x-10}{60}\right),$

first simplify to $$3\left(\dfrac{x^2-15x+50}{28}\right) + 12\left(\dfrac{x^2-8x-20}{35}\right) + 11\left(\dfrac{x^2-3x-10}{20}\right).$$

Now give the fractions their lowest common denominator:

$$15\left(\dfrac{x^2-15x+50}{140}\right) + 48\left(\dfrac{x^2-8x-20}{140}\right) + 77\left(\dfrac{x^2-3x-10}{140}\right).$$

Now the fractions can be added: $$\dfrac{15\left({x^2-15x+50}\right) + 48\left({x^2-8x-20}\right) + 77\left({x^2-3x-10}\right)}{140}.$$

Can you conclude?

  • $\begingroup$ $140= $ lcm $ (28, 35, 20)$ $\endgroup$ – J. W. Tanner May 1 at 17:10
  • $\begingroup$ Sorry I am still a bit confused with how you add those fractions with a multiply outside still. Would you combine the top parts and have it over 140 and then multiply by the 15*48*77? $\endgroup$ – perkss May 1 at 17:20
  • $\begingroup$ I edited my answer to help with your additional question; you could let me know if you need further help $\endgroup$ – J. W. Tanner May 1 at 17:27
  • $\begingroup$ Thank you so much got it worked out now! $\endgroup$ – perkss May 1 at 17:32
  • $\begingroup$ You're welcome. I'm glad $\endgroup$ – J. W. Tanner May 1 at 17:34

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