Answer:
#1
Well lets solve the polynomial first.
(4p +3) ^2
That is basically
(4p + 3) (4p + 3)
And we’re trying to find the error in the “equal” solution.
16p^2 + 8p + 9.
So let’s solve the polynomial first
Following F.O.I.L
F- First
O- Outter
I- Inner
L- Last
So doing the equation in order.
4p * 4p = 16p^2
4p * 3 = 12p
3 * 4p = 12p
3* 3 = 9
Added it all up.
16p^2 + 12p + 12p + 9
Simplify.
16p^2 +24p +9
We can see that in this case the 8p is wrong.
So we replace 8p with 24p
In this case the last option.
Factoring #2
So we’re factoring
9x^2 -6x - 3
So how are we gonna do this.
So I’m just gonna go over the simple method.
The standard form of a quadratic is
Ax^2 + bx + c = 0
We don’t have to worry about the equal 0 part here for now.
ax^2 + bx + c
A being the number next to x^2
B is the number being next to x
C is the number that is next to nothing, jsut a plain whole number (sometimes whole number)
So let’s plug it in.
(It’s already plugged in so we can solve)
9x^2 - 6x - 3
Multiply a and c in ths case being, 9 and -3.
You’d get -27
We need two numbers that multiply to -27 and add up to -6
From here it’s guessing and checking.
Here’s my process.
trial 1
Since the sum is negative, it has to be negative + negative or positive + negative greater then the positive. Or vice versa but positive is less then negative.
But the product has to be -27,negative * negtaive is positive, so it has to be positive * negative and positive + negative.
I look at -27 and think of two numbers that would give the lowest.
-9 * 3
-9 * 3 = -27
-9 + 3 = -6
It looks like I got it already on the first try.
The numbers are -9 and 3
end of trial 1
So now I know that the numbers are -9 and 3.
Using that information I just plug it into the quadratic but replace it with b.
9x^2 -9x + 3x - 3
Now you can simplify this
In the end you’d get
3 (3x+1)(x-1)
You’d have the third option open to you
Figure out the second one using these strategies
