Question

The expression $21x^2+ax+21$ is to be factored into two linear prime binomial factors with integer coefficients. This can be done if a is:

• A. odd number
• B. zero
• C. even number
• D. None

Answer 1

Answer: (C)

even number

Let the factors be $Ax+B$ and $Cx+D$. Then
$(Ax+B)(Cx+D)=AC{x}^2+(AD+BC)x+BD=21x^2+ax+21$.
$\therefore AC=21,BD=21$. Since 21 is odd, all its factors are odd. Therefore each of the numbers $A$ and $C$ is odd. The same is true for $B$ and $D$. Since the product of two odd numbers is odd, $AD$ and $BC$ are odd numbers;
$a=AD+BC$, the sum of two odd numbers, is even.