Part A. Lest fin the zeros of our function first:
[tex]f(x)=-x^2-7x+30[/tex]
Step 1. Set the right side of the function equal to 0:
[tex]0=-x^2-7x+30[/tex]
[tex]-x^2-7x+30=0[/tex]
Step 2. Multiply both sides of the equation by -1
[tex](-1)(-x^2-7x+30)=0(-1)[/tex]
[tex]x^2+7x-30=0[/tex]
Step 3. Factor the left side:
[tex](x+10)(x-3)=0[/tex]
Step 4. Set each factor equal to 0 and solve for [tex]x[/tex]:
[tex]x+10=0[/tex] and [tex]x-3=0[/tex]
[tex]x=-10[/tex] and [tex]x=3[/tex]
Since distances can't be negative, the zero of our function is [tex]x=3[/tex].
We can conclude that the zero [tex]x=3[/tex] represents that when the horizontal distance of the diver is 3 ft, he is is in the surface of the pool. In other words, his vertical distance is zero.
Part B. To factor our quadratic equation [tex]9x^2+3x-2=0[/tex], we are going to use the method for simple cases because [tex]a \neq 0[/tex]:
Step 1. Find tow numbers whose product is [tex]ac[/tex] and its sum is [tex]b[/tex]:
We know form our quadratic that [tex]a=9[/tex], [tex]b=3[/tex], and [tex]c=-2[/tex], so [tex]ac=9(-2)=-18[/tex].
We want tow numbers that multiply together to make -18, and add up to 3.
Those number are 6 and -3. [tex](6)(-3)=-18[/tex] and [tex]6-3=3[/tex]
Step 2. Rewrite the middle term [tex]3x[/tex] using those numbers:
[tex]9x^2+6x-3x-2=0[/tex]
Step 3. Factor the expression:
[tex](9x^2+6x)-(3x+2)=0[/tex]
[tex]3x(3x+2)-(3x+2)=0[/tex]
[tex](3x+2)(3x-1)=0[/tex]
