Question

. (x – 1) is a factor of p(x) if and only if the sum of coefficients of p(x) is _​

Answer 1

Answer:

x=1 is a factor, the sum of coefficients of p(x) is 0.

Step-by-step explanation:

From the above question,

They have given :

(x – 1) is a factor of p(x) if and only if the sum of coefficients of p(x) is _​

x−1  is a factor of a polynomial P of positive degree if and only if the sum of the coefficients of P is zero.

If x−1x−1 is a factor then you have p(x)=(x−1)q(x)p(x)=(x−1)q(x) where q(x) $ is another polynomial.

Let x=1x=1, we get

p(1)=(1−1)q(1)=0p(1)=(1−1)q(1)=0

That is if you let x=1x=1 in your polynomial you will get p(1)=0p(1)=0

Note that when you find p(1)p(1) you let x=1x=1 , so you are just adding the coefficients of P(x)P(x) together.

Thus if x=1x=1 is a factor, the sum of coefficients of p(x)p(x) is 00.

The remainder of the division of a polynomial p(x)p(x) by x−ax−a is p(a)p(a), so

p(x) divisible by x−a⟺p(a)=0.p(x) divisible by x−a⟺p(a)=0.

On the other hand p(1)p(1) is just the sum of the coefficients of p(x)p(x),

Thus, if x=1 is a factor, the sum of coefficients of p(x) is 0.

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Answer 2

Answer:

$\qquad\qquad\boxed { \sf \: Sum\:of\:coefficient \: of \: p(x) \: = \: 0}$

Step-by-step explanation:

Let us consider a polynomial of degree n as

$\sf \: p(x) = a_0 {x}^{n} + a_1 {x}^{n - 1} + a_2 {x}^{n - 2} + ... + a_n$

Now, it is given that,

$\sf \: x - 1 \: is \: a \: factor \: of \: p(x)$

By using factor theorem, we have

$\sf \: p(1) = 0$

$\sf \: a_0 {(1)}^{n} + a_1 {(1)}^{n - 1} + a_2 {(1)}^{n - 2} + ... + a_n = 0$

$\sf\implies \sf \: a_0 + a_1 + a_2 + ... + a_n = 0$

$\sf\implies \sf \: Sum\:of\:coefficient \: of \: p(x) \: = \: 0$

$\rule{190pt}{2pt}$

Factor Theorem :- This theorem states that if x - a is a factor of polynomial f(x) of degree greater than or equals to one, then f(a) = 0

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$