Question

SOMEBODY PLZ!! ITS THE LAST DAY I DONT WANNA FAIL :*(

Answer 1

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