Answer:
2, -4, -10 and -18.
Step-by-step explanation:
The given expression is
\(x^2-3x+b\) ...(i)
We need to find the 4 values of b which make the expression factorable.
A polynomial is factorable if both roots are real.
If \(\alpha \text{ and }\beta\) are two real roots of a polynomial, then the polynomial is defined as
\(P(x)=x^2-(\alpha+\beta)x+\alpha\beta\) ....(ii)
From (i) and (ii), we get
\(\alpha+\beta=3\) ...(iii)
\(\alpha\beta=b\)
For equation (iii), possible pairs of \(\alpha \text{ and }\beta\) are (2,1), (4,-1), (5,-2) and (6,-3).
From these ordered pairs the values of b are
\(b=\alpha\beta=2\times 1=2\)
\(b=\alpha\beta=4\times (-1)=-4\)
\(b=\alpha\beta=5\times (-2)=-10\)
\(b=\alpha\beta=6\times (-3)=-18\)
Therefore, the four possible values of b are 2, -4, -10 and -18.
