Question 162·Hard·Advanced Math

Try $x=1$ because every power of $1$ stays $1$, making the terms quick to test. Type $p(x)=x^4-10x^2+9$, then $p(1)$ in Desmos. It prints $0$. At an input where one form is zero, the other form must be zero too.

The problem says the product is the same polynomial, so it must also equal $0$ at $x=1$. Substitute $1$ for each $x$ and combine the constants:

Type $(4-a)(4+a)\sim p(1)\{a>0\}$ in Desmos. The $\sim$ asks Desmos to fit $a$ to the equation, and the braces enforce the positive condition. Under PARAMETERS, Desmos gives $a=4$. So the positive constant in the factorization is $4$. Choice C.

The factors have the same $x^2+3$ part, but opposite $ax$ parts. That’s a difference of squares: $(u-v)(u+v)=u^2-v^2$. Rewrite the product:

The opposite terms cancel instead of producing an $x^3$ term.

Expand the square, then combine the like terms, the terms that both contain $x^2$:

So $6-a^2$ is the coefficient multiplying $x^2$.

The two forms are the same polynomial, so their $x^2$ coefficients must agree. The given polynomial has $-10x^2$; match its coefficient to the one you found:

Subtract $6$:

Divide by $-1$:

Take the positive square root, since the problem says $a$ is positive:

So the constant in the factorization is $4$. Choice C.