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