After factoring out the common binomial, you will subtract one binomial from the other. Write that subtraction explicitly (for example, something like $(2x-5)-(x-2)$) and be careful with the minus signs when you simplify.

Once you have the expression in the form $ax^2+bx+c$, remember that $a+b+c$ is just the sum of the three coefficients. As a shortcut, note that for any polynomial $ax^2+bx+c$, the value when $x=1$ is equal to $a+b+c$. You can use this to check your work.

In Desmos, type

f(x) = (2x-5)(x+3) - (x-2)(x+3)

on one line. Desmos will graph the corresponding parabola for this function.

On a new line in Desmos, type f(1). The numerical value Desmos shows is the value of the quadratic when $x=1$, which is equal to $a+b+c$ for the polynomial $ax^2+bx+c$.

Start with the expression:

Both terms contain the factor $(x+3)$, so factor it out using the distributive property $AB-AC=A(B-C)$:

Now simplify the expression inside the brackets:

So the whole expression becomes

Multiply $(x+3)(x-3)$ to get the quadratic in standard form:

So the original expression can be written as

which matches the form $ax^2+bx+c$.

From $x^2-9$ we have:

• $a=1$ (coefficient of $x^2$),
• $b=0$ (there is no $x$ term),
• $c=-9$ (constant term).

Now compute

So the value of $a+b+c$ is $-8$.