Simplify $9((x^2-15x+50)/84)-12((x^2-8x-20)/-35)+33((x^2-3x-10/60)$

Hi I am trying to simplify the following I found online

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

to

$$= x^2 -6x -7$$

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.

• Welcome to Mathematics Stack Exchange. You don't want to multiply by $84$; you want to divide by $84$. Try finding a common denominator – J. W. Tanner May 1 at 17:00
• 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. – 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?

• $140=$ lcm $(28, 35, 20)$ – J. W. Tanner May 1 at 17:10
• 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? – perkss May 1 at 17:20
• I edited my answer to help with your additional question; you could let me know if you need further help – J. W. Tanner May 1 at 17:27
• Thank you so much got it worked out now! – perkss May 1 at 17:32
• You're welcome. I'm glad – J. W. Tanner May 1 at 17:34