How to do Synthetic Division by 2x-1 | VCE METHODS

[@milavabrudi3680]

This was genuinely a life saver🙏🙏Appreciate this forreal

[@arlenemagpantay8386]

Why is the last number didn't change when you need to divide it by 2? But the temporary divisor goes back to it's original divisor?

[@MathBase]

If you're referring to the 3/16(x-1/2) bit at the end, then it's because only the 16 OR the (x-1/2) can multiply 2. But not both. When purely viewing the denominator element, it's essentially 2 * 16 * (x-1/2). The 2 can only multiply one of the factors following it; 16 or (x-1/2). It's cleaner to multiply the (x-1/2) to eliminate that small fraction so that's what I did. Hope that makes sense! :)

[@sanguine9031]

Man u the goat man i cant lie

[@MathBase]

Thanks!

[@ygfrt21]

I have a test next set thanks

[@bami_codes]

Well explained.

[@MathBase]

Thank you so much!

[@angel_5606]

Thanks

[@danielnicholson2949]

Thank you

[@MathBase]

You're welcome!

[@asu.ra21]

Wait, so example, if your answer is a whole number (3x^2, 5x, 1, 8) but if your given is still 2x + 3, do i still divide it or do i convert the 2 to be multiplication?

[@asu.ra21]

Cause it's my first time to encounter a non-monic synthetic div. so sorry if I dont get it in the first time

[@MathBase]

@@asu.ra21 All good. Non-monics are tough to understand. The short version is you always divide the number. So divide the final answer using the 2.

[@diliseekers]

So, please, what is the remainder after dividing?

[@MathBase]

For the final question, the remainder function is 3/16(2x-1). The actual remainder is just 3/16. :)

[@soso4720]

thank you very much

[@MathBase]

I'm glad this helped!

[@Lex54321]

So it means I can divide my answer, either 2 or 1/2?

[@MathBase]

It means when the polynomial divides 2x-1, first synthetic divide by 1/2. Then, take what you found and divide 2. So you must do both divisions.
Another example, if the polynomial divides 4x+3, first synthetic divide -3/4, then divide 4. Both must be done. Hope that makes sense! :)

[@OmKarmacharya-fi5fx]

Well explained😊

[@bobabooks]

thanks!🤭

[@MathBase]

You're welcome!

[@abelotube5864]

Nice 👍

[@MathBase]

Thank you!