Question

find all integers k such that trinimial can be factored over the integers 3x squared +kx-5

Answer 1

The question breaks down like this:
$$
~~~~~~3x^2+kx-5\\
=\frac{( 3x+a )(3x+b))}{3}\\
=\frac{9x^2+3x\times b+ax\times3+a\times b}{3}\\
=3x^2+bx+ax+\frac{a\times b}{3}\\
=3x^2+x(b+a)+\frac{a\times b}{3}\\
=3x^2+kx-5
$$
This can be true only if \(\dfrac{a\times b}{3}=-5\), which is that same as \(a\times b=-15\)
We need to find all factors of -15.
These would be \( \{a,b\}=\{-1,15\},\{1,-15\},\{3,-5\},\{-3,5\} \). These are the only possibilities for \(\{a,b\}\).

The next requirement is that \( a+b=k\).

Using only these factors of -15, what can k be?

Answer 2

if you know the AC method of factoring,
rewrite \[3x^2+kx-5\] as \[x^2+kx-15\]
now find all possible factors of 15 and add them, giving you all the possible values of k.

Answer 3

correction. ... all possible factors of (-15) ...