P(x) = (x { }^(2) - 1(x { }^(2) - 5x + 6)Can you put this in a better understanding for me please?

Question

`p(x) = (x { }^(2) - 1(x { }^(2) - 5x + 6)`Can you put this in a better understanding for me please?

Answer 1

`\begin{gathered} (x^2-2)(x^2-5x+6)=0 \\ we\text{ eqaute this to find the zeroes} \end{gathered}`

`\begin{gathered} (x^2-3x)(-2x+6)=0,\text{ so we group th}em\text{ after replacing -5x by -3x and -2x},\text{ we factorize} \\ x(x-3)\text{ -2(x-3)}=0 \\ (x-2)(x-3)=0 \\ x=2 \\ or\text{ } \\ x=3 \end{gathered}`

values of the zeroes of the function are

`\begin{gathered} x=\text{ +}\sqrt[]{2} \\ x=-\sqrt[]{2}_{} \\ x=\text{ 2} \\ x=\text{ 3} \\ \text{Factors are (x-}\sqrt[]{2}),\text{ (x+}\sqrt[]{2}),(x-3)and\text{ (x-2)} \end{gathered}`