Question

Consider the polynomial $P\left(x\right)=x^2-6x+9$
(a) Find $P\left(3\right)$
(b) Prove that the value of this polynomial cannot be negative numbers.
(c) Find two numbers $a$ and $b$ such that $P\left(a\right)=P\left(b\right)$.

Answer 1

(a) $P\left(x\right)=x^2-6x+9$
$\Rightarrow$ $P\left(3\right)=3^2-6\left(3\right)+9=9-18+9=0$
(b) $P\left(x\right)=x^2-6x+9=x^2-6x+9=-2\times 3\times x+3^2={(x-3)}^2$, which is a perfect square, therefore the number will always be positive
Hence, the value of this polynomial cannot be a negative number.
(c) $P\left(a\right)=a^2-6a+9$
$P\left(b\right)=b^2-6b+9$
$P\left(a\right)+P\left(b\right)$
$\Rightarrow a^2-6a+9=b^2-6b+9$
$\Rightarrow a^2-b^2-6a+6b=0$
$\Rightarrow (a^2-b^2)-6(a-b)=0$
$\Rightarrow (a-b)(a+b)-6(a-b)=0$
$\Rightarrow (a-b)(a+b-6)=0$
$\Rightarrow (a-b)=0\Rightarrow a=b$ (Not possible)
$a+b=6$
There can be many numbers as long it satisfies $a+b=6$
For example, $a=1,b=5$