Question

P(x) is a second degree polynomial with P(1)=0 and P(-2)=0.
a) Find two first degree factors of P(x).
b) Find the polynomial P(x).

Answer 1

Hint: Given polynomial is a second degree polynomial. So, there will be two roots as it is of second degree. If $f\left( x \right)$ is a polynomial and f(x)=0 at a certain x, then x is a root for that polynomial. Here at x=1, x=2 the value of the polynomial is zero. Using this, build the polynomial P(x)

Complete step-by-step answer:
We are given a polynomial P(x) is of second degree and the value of P(1) is zero and the value of P(-2) is zero.
We have to find the polynomial P(x) and two first degree factors for the polynomial P(x).
We already know that, according to the Factor theorem, if $f\left( x \right)$ is a polynomial and $f\left( a \right) = 0$, then $x - a$ is a factor of $f\left( x \right)$.
Here it is given that $P\left( x \right)$ is a polynomial and P(1)=0
This means $x - 1$ is a factor of $P\left( x \right)$
It is already given that P(-2) is also equal to zero.
This means $x - \left( { - 2} \right) = x + 2$ is also a factor of $P\left( x \right)$
As $P\left( x \right)$ is a second degree polynomial, it can have only two factors.
We already got two factors which are $x - 1,x + 2$
So, the two first degree factors of $P\left( x \right)$ are $x - 1,x + 2$
A polynomial is the product of its factors
$P\left( x \right)$ is the product of $x - 1$ and $x + 2$
$
  P\left( x \right) = \left( {x - 1} \right)\left( {x + 2} \right) \\
  \to P\left( x \right) = x\left( {x + 2} \right) - 1\left( {x + 2} \right) \\
  \to P\left( x \right) = {x^2} + 2x - x - 2 \\
  \therefore P\left( x \right) = {x^2} + x - 2 \\
$
Therefore, the polynomial $P\left( x \right)$ is ${x^2} + x - 2$

Note: Another approach to find the polynomial
General form of a polynomial of second degree (quadratic polynomial) is ${x^2} - \left( {a + b} \right)x + ab$, where a and b are the roots of the polynomial.
Here we already got that $x - 1,x + 2$ are the factors of the polynomial $P\left( x \right)$ which means 1 and -2 are the roots of the polynomial. Substitute the roots in the general form of the polynomial to obtain the polynomial.
$
  a = 1,b = - 2 \\
  P\left( x \right) = {x^2} - \left( {a + b} \right)x + ab \\
 \to P\left( x \right) = {x^2} - \left( {1 + \left( { - 2} \right)} \right)x + \left( {1 \times - 2} \right) \\
  \to P\left( x \right) = {x^2} - \left( { - 1} \right)x + \left( { - 2} \right) \\
  \therefore P\left( x \right) = {x^2} + x - 2 \\
$