The given function is
[tex]f(x)=x^2+3x-10[/tex]
We need to find the solution of
[tex]\begin{gathered} f(x)\leq0 \\ x^2+3x-10\leq0 \end{gathered}[/tex]
Since the sign of inequality is <=, then the solution will be closed intervales from the small number to the big number [a, b]
At first, factor the left side into 2 factors
[tex]\begin{gathered} x^2=(x)(x) \\ -10=(-2)(5) \\ (x)(-2)+(x)(5)=-2x+5x=3x \end{gathered}[/tex]
Then the two factors are (x - 2) and (x + 5)
[tex]\begin{gathered} x^2+3x-10=(x-2)(x+5) \\ (x-2)(x+5)\leq0 \end{gathered}[/tex]
Now, equate each factor by 0 to find the values of x
[tex]\begin{gathered} x-2=0,x+5=0 \\ x-2+2=0+2,x+5-5=0-5 \\ x=2,x=-5 \end{gathered}[/tex]
Since the smallest number -s -5 and the greatest number is 2, then
[tex]-5\leq x\leq2[/tex]
The solution should be a closed interval from -5 to 2
The solution is [-5, 2]
The answer is
A. The solution is [-5, 2]
