Question

find the value of p for which x+p is a factor of x square+px+3-p

Answer 1

The value of "p is equal to 3".

Step-by-step explanation:

Let the given polynomial $P(x):x^2+px+(3-p)$

To find, the value of p = ?

(x + p) is a factor of P(x).

∴ x + p = 0

⇒ x =  - p

∴ $P(x):x^2+px+(3-p)=0$

$x^2+px+(3-p)=0$     ...... (1)

Put x = = - p in (1), we get

$(-p)^2+p(-p)+3-p=0$

⇒ $p^2-p^{2} +3-p=0$

⇒ $3-p=0$

⇒ p = 3

Hence, the value of "p is equal to 3".

Answer 2

The value of p is 3.

Step-by-step explanation:

Here, the given polynomial is

$\quad f(x)=x^{2}+px+3-p$

Since $(x+p)$ is a factor of $f(x)$, then we can conclude that $x=-p$ is a zero of $f(x)$.

Then, $f(-p)=0$

$\Rightarrow (-p)^{2}+p(-p)+3-p=0$

$\Rightarrow p^{2}-p^{2}+3-p=0$

$\Rightarrow 3-p=0$

$\Rightarrow p-3=0$

$\Rightarrow \boxed{p=3}$

Check step:

Let us check whether $x+p=x+3$ is a factor of $f(x)$ or not.

When $p=3$,

$\quad f(x)=x^{2}+3x+3-3$

$\Rightarrow f(x)=x^{2}+3x$

$\Rightarrow f(x)=x(x+3)$

This shows that $(x+3)$ is a factor of $f(x)$.