Respuesta :
The Solution To Question Number 10:
The question says what operations would the set of quadratics be closed under.
Let the sets of quadratics be
[tex]\begin{gathered} p(x)=ax^2+bx+c \\ q(x)=mx^2+nx+k \end{gathered}[/tex]The set of two quadratics (polynomials) is closed under Addition.
Explanation:
[tex]\begin{gathered} P(x)+q(x)=(ax^2+bx+c)+(mx^2+nx+k) \\ =(a+m)x^2+(b+n)x+(c+k) \\ \text{which is still a quadratic.} \\ \text{Hence, the set of quadratics is closed under Addition.} \end{gathered}[/tex]The set of two quadratics is closed under Subtraction.
[tex]\begin{gathered} P(x)-q(x)=(ax^2+bx+c)-(mx^2+nx+k) \\ =(a-m)x^2+(b-n)x+(c-k) \\ \text{which is still a quadratic, provided both a}\ne m,\text{ b}\ne n\text{ } \\ \text{Hence, the set of quadratics is closed under Subtraction.} \end{gathered}[/tex]The set of quadratics is not closed under Multiplication.
[tex]\begin{gathered} P(x)\text{.q(x)}=(ax^2+bx+c)(mx^2+nx+k)=amx^4+(bn+ak)x^2+ck+\cdots \\ \text{Which is not a quadratic.} \\ \text{Hence, the set of quadratics is not closed under multiplication.} \end{gathered}[/tex]The set of quadratics is not closed under Division.
[tex]\begin{gathered} \text{Let the sets be f(x)=8x}^2\text{ and} \\ h(x)=2x^2-1 \\ \text{ So,} \\ \frac{f(x)}{h(x)}=\frac{8x^2}{2x^2_{}-1} \\ \text{Which is not a quadratic.} \\ \text{Hence, the set is not closed under Division.} \end{gathered}[/tex]