Find the sum of coefficients

The sum of the coefficients of a polynomial, p(x) = p(1). You may think of it as the remainder when the polynomial is divided by x-1. I did not provide a proof in this video since the purpose is to make the principle known.

[benmooiman]

I’m 68 years old now and i enjoy every video 👍 “never stop learning …” 🇳🇱

[auztenz]

Seems logical
To prove ot we first need to define what the sum of the coefficient is
Let p(x) = a_nx^m_n+a_(n-1) x^....
m1<m2<m3... <m_n
Then the sum of the coefficient is
A_n+a_(n-1)+a_(n-2) ...a1 + constant
= a_n×1+a_(n-1) ×1...
When x=1
a_n×1 +a_(n-1) ×1 +...
Which is the same as the sum of the coefficient
Hence proved

[BobChao]

I exclaimed out loud when you said what the trick was! It's so obvious, so beautiful, so simple! It's almost what I did, but because I didn't phrase it as P(1), I dismissed it. Thank you for this cool trick ☺️

[mmfpv]

Nice, thanks! Good to have that distilled into a practical concept. FYI, the preview picture for your video shows a slightly different problem with the x^3 coefficient being 3, but then in problem you solve it is 31. No bother though, the concept comes through

[antonionavarro]

Excellent. I always learn something new.

[Grecks]

For any polynomial, we get the sum of all coefficients if we evaluate the polynomial function at x=1, right? P(1) = -32 in this case, so -32 should be the answer.

[iMustBro]

I really love your videos professor.

Thanks for reigniting my interest in Math!

[SIDDHARTHSINGH-ppuz]

2nd method if we need sum of coefficients of
Even place than put
(x=1) - (x=-1) whole ÷2
Odd place than put
(x=1) + (x=-1) whole ÷2
And for sum of coefficients
Add even place term and odd term place .

Sorry if anywhere language mistake

[Michaeladjei]

Your videos give me joy 😊❤
Whenever I end one, I look for more. Keep it up sir 😅

[bhargavkulkarni]

Your videos take me back to my 12th standard days. Days of my unrelenting, rigorous mathematics practice

[hollyrockerzzz]

Dear I have a problem in a question related to time and distance can u plz solve that and where can I send u that question

[UGF]

The trick to finding the sum of all the coefficients of a polynomial raised to a power is to simply plug in 1 for the variable.
When a polynomial is raised to a power, the sum of the coefficients in the expanded polynomial is equal to the value of the polynomial when 1 is substituted for the variable.
The process of expanding the polynomial and adding up all the coefficients can be skipped by using the shortcut of plugging in 1 for the variable.
The example of squaring the polynomial 2x + 1 demonstrates how the sum of the coefficients is equal to the value of the polynomial when 1 is substituted.
Applying the trick of plugging in 1 is useful when solving this type of problem, especially in a competition setting where time is limited.

[shravan]

But isnt 12 not a coefficient are we supposed to consider that in the sum? Maybe cuz u can write it as 12 times x^0 ?. I did come to the same conclusion as finding p(1) but got confused here.

[wyojohn]

Just another way of showing how it works. (2x + 1)^2 = (2x)(2x) + (2x)1 + 1(2x) + 1(1) = (2)(2)(x)(x) + 1(2)x + 1(2)x + 1(1). Thus the coefficients would be (2)(2), 1(2), 1(2) and 1(1). And you can see that if you just let x=1, then (2)(2)(x)(x) + 1(2)x + 1(2)x + 1(1) = (2)(2) + 1(2) + 1(2) + 1(1), which is the sum of the coefficients.

[davidplanet]

I misunderstood the question and took derivatives and set x=0 to get the coefficients. Then I added them. A long way to get the same answer.

[maxhagenauer]

Actually the question was the sum of coefficients AFTER you expand out so you had to solve it by expanding out.

[ZoonCrypticon]

The question would be, "what is the use of the sum of the coefficients" ? Does it have any practical application ?