Math trinom help

Question

$9x^2-9$ its like $(3x+3) (3x-3)$ what about $9x^2-35$ ?

Answer 1

$$9x^2 - 35 = \left(3x\right)^2-\left(\sqrt{35}\right)^2 = \left(3x+\sqrt{35}\right)\left(3x-\sqrt{35}\right)$$

Answer 2

In general $$ \forall a,b \in \mathbf{R}, \,\,a^2-b^2=(a+b)(a-b) $$

Answer 3

If you forgot what the difference of squares (or any other terms related to factorization) is and just want to factorize a polynomial expression, Check this wiki page out: https://en.wikipedia.org/wiki/Discriminant

In general, polynomials of second order (meaning the highest power of x is 2) have a discriminant that can be computed. The Discriminant is used to factorize the polynomial expression. For a given second order polynomial $p(x) = ax^2+ bx + c$, the formula for the discriminant is $\Delta = b^2 - 4ac$.

After finding $\Delta$, you will be able to find the roots ($x_{1}$ and $x_{2}$) of the polynomial and factorize it. After factorization, $p(x)$ will be $p(x) = a(x-x_{1})(x-x_{2})$

In your case, your polynomial expression is $p(x) = 9x^2 - 35 = 9x^2 +0x -35$
From here, $a=9, b=0, c=-35$. With this,

$\Delta = 0^2 - 4(9)(-35) = 1260$.

Now you can use $\Delta$ to find the roots of the polynomial. For a second order polynomial, there are at most 2 roots.

$x_{1} = \frac{-b-\sqrt{\Delta}}{2a}$, $x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

From here, $x_{1}=\frac{-\sqrt{1260}}{18}=\frac{-\sqrt{35}}{3}, x_{2}=\frac{\sqrt{1260}}{18}=\frac{\sqrt{35}}{3}$

Since we have the roots, we can express $p(x)$ in its factorized form:
$p(x) = 9(x-\frac{-\sqrt{35}}{3})(x-\frac{\sqrt{35}}{3})=(3x+\sqrt{35})(3x-\sqrt{35})$

This general method should allow you to feel comfortable with any second order polynomial; therefore, you won't have to remember some special cases.