Question

Bill wants to factor x^2+8x+16 by grouping; however sara says it is a special product and can factor a different way. using complete sentences, explain and demonstrate how both methods will result in the same factors.

Answer 1

It's a perfect square, I would just factor by inspection. That could be one method, lol

Answer 2

I suck at math so that makes no sense.

Answer 3

(a+b)^2=a^2+2ab+b^2

Answer 4

Hm. So you know that since there's no coefficient before x^2 your factor starts with

(x

and the number(s) following need to multiply by to give you 16 and add to give you 8--and 4 work, so it's a perfect square.

Answer 5

okie so what does it mean by sara saying it's a special product?

Answer 6

It's special because your factors are the same. :)

Answer 7

So what am I suppose to say??

Answer 8

I don't understand what this question wants me to do!!!

Answer 9

Please somebody help me???

Answer 10

So they want you to use two different methods.
Let's look at Bill's method first (grouping).\[\Large\bf\sf x^2+8x+16\quad=\quad x^2+4x+4x+16\]8x can be written as 4x and 4x.
From here we'll factor an x out of each of the first two terms,\[\Large\bf\sf x(x+4)+4x+16\]And factor a 4 out of each of the last two terms,\[\Large\bf\sf x(x+4)+4(x+4)\]

Answer 11

From there we can see that they each have a factor of (x+4), so we'll factor that out of each,
\[\Large\bf\sf \color{orangered}{x}(x+4)\color{orangered}{+4}(x+4)\quad=\quad (x+4)(\color{orangered}{x+4})\]

Answer 12

The same factor is being multiplied by itself, so we can write it as a square,\[\Large\bf\sf (x+4)(x+4)\quad=\quad (x+4)^2\]

Answer 13

So that's Bill's method.
Make sensssseeeee, kinda?

Answer 14

Yeah. kind of. math is like learning a very old language to me at times.
So what would sara's method be?

Answer 15

Sara remembers that there is a special way to simplify the `sum of squares`.
\[\Large\bf\sf a^2+2ab+b^2\quad=\quad (a+b)^2\]So we start with:\[\Large\bf\sf x^2+8x+16\]We need to write 16 as a square,
and also need to write 8x as `2 times something`. And hopefully that something will be our a and b.

Answer 16

16 is 4^2 right?
And 8x is 2*4x.

Answer 17

\[\Large\bf\sf x^2+8x+16\quad=\quad x^2+2(4x)+4^2\]

Answer 18

See how we can use the rule for the `sum of squares` to simplify it?

Answer 19

\[\Large\bf\sf a^2+2(ab)+b^2\quad=\quad (a+b)^2\]\[\Large\bf\sf x^2+2(4x)+4^2\quad = \quad ?\]

Answer 20

Sorry... but I am totally lost. In math... I feel like I should be in 5th grade instead of 12th!!

Answer 21

:c

Answer 22

srry

Answer 23

This is the rule for the sum of squares:\[\Large\bf\sf \color{royalblue}{a}^2+2(\color{royalblue}{a}\color{orangered}{b})+\color{orangered}{b}^2\quad=\quad (\color{royalblue}{a}+\color{orangered}{b})^2\]
Understand how we can apply it to our problem here?
\[\Large\bf\sf \color{royalblue}{x}^2+2(\color{orangered}{4}\color{royalblue}{x})+\color{orangered}{4}^2\quad=\quad ?\]
We're trying to turn it into something like the rule shows on the right.

Answer 24

so it's (x+4)^2

Answer 25

Yes, good.

Answer 26

so what do I do next?

Answer 27

Both Bill and Sara's methods resulted in (x+4)^2, the same answer.

Explain in sentences how you got the same answer using different methods.
Use my steps to try and explain it.

Answer 28

How is this???

Bill got (x+4)^2 by doing x^2+4x+4x+16. He than factored out an x from each of the first two terms, getting x(x+4) + 4x+16. Next he factored out a 4 out of each of the last two terms, x(x+4) + 4(x+4). He than factored out (x+4), since they each have it. Bill then got (x+4)(x+4), which he then squared it being that the same factor was being multiplied by itself. He ends the problem with (x+4)^2.

Sara knows that a^2+2ab+b^2= (a+b)^2. So Sara starts with x^2+8x+16, and knows that she needs to write 16 as a square and 8x will be 2 times 4x. So Sara gets, x^2+2(4x)+4^2. Being that the rule for sum of squares is a^2+2(ab)+b^2=(a+b)^2, Sara knows that the x in X^2 is the a and the 4s in 2(4x)+4^2 is the b's, she gets (x+4)^2.

Answer 29

lol yayyyy good job! \c:/

Answer 30

So that makes sense??

Answer 31

Just seems like you copied what I wrote lolol XD
Yes it makes sense, I'm hoping it makes at least a little sense to you :3

Answer 32

Yeah going throught it helped.

Answer 33

Can you help me with this question?
Three functions are given below: f(x), g(x), and h(x). Explain how to find the axis of symmetry for each function and rank the functions based on their axis of symmetry (from smallest to largest). (10 points)
f(x) = 3(x + 4)2+ 1
g(x) = 2x2 -16x + 15
h(x) is graphed.

Answer 34

So if you look at the graph, where is the line of symmetry for the curve drawn?

In other words: Where could we fold the paper in half to make the curve overlap itself perfectly.

Answer 35

(1,-3)

Answer 36

Ok good that's where the vertex of the parabola is located.
But if want an axis of symmetry then we need a line that we can fold over.

So we want that entire line going up, x=1.
See how the graph is mirrored on both sides of that line?

Answer 37

yeah.

Answer 38

Notice how the axis of symmetry was simply the `x coordinate of our vertex`?
The other functions will work out just as easily.

Do you know how to find the vertex of f(x) and g(x)?

Answer 39

I understand this question thanks. I have had the following question for a while...
Sandra exclaims that her quadratic with a discriminant of -4 has no real solutions. Sandra then puts down her pencil and refuses to do any more work. Create an equation with a negative discriminant. Then explain to Sandra, in calm and complete sentences, how to find the solutions, even though they are not real. (10 points)

Answer 40

Woah calm down Sandra! Let's find the complex solutions as well! :U

Answer 41

So we need to start with an equation that has a negative discriminant?
Umm sec lemme think of something.

Answer 42

\[\Large\bf\sf x^2-2x+5\]

If we throw this polynomial into the Quadratic Formula:\[\Large\bf\sf x\quad=\quad\frac{-b\pm \sqrt{b^2-4ac}}{2a}\quad=\quad \frac{2\pm\sqrt{4-4\cdot1\cdot5}}{2(1)}\]

Answer 43

Our discriminant is the stuff under the radical.

\(\Large\bf\sf 4-4\cdot5\quad=\quad 4-20\quad=\quad -16\)
Negative is bad when under root, yes?

Answer 44

um.... yes?

Answer 45

Yes. We can't take the square root of a negative number. Bad.
But we have a way of classifying these bad numbers by calling them "imaginary".

The way we do this is by calling \(\Large\bf\sf \sqrt{-1}=\mathcal i\) .

Answer 46

Here is a way we can deal with the negative under the square root,\[\Large\bf\sf \sqrt{-16}\quad=\quad \sqrt{-1\cdot 16}\quad=\quad \sqrt{-1}\cdot \sqrt{16}\]See how we were able to separate the bad number from the rest of our stuff?

Answer 47

\[\Large\bf\sf \sqrt{-1}\cdot\sqrt{16}\quad=\quad \mathcal i\cdot \sqrt{16}\]

Answer 48

Applying the Quadratic Formula to our quadratic function we get,\[\Large\bf\sf x\quad=\quad \frac{2\pm\sqrt{16}\cdot \mathcal i}{2}\]Which can be simplified down a bit.
I think 16 is a perfect square isn't it?
Can we take it out of the root somehow?

Answer 49

I don't think so??

Answer 50

lol :)

Answer 51

wat?

Answer 52

\[\Large\bf\sf 4\cdot 4\quad=\quad 16\]

\[\Large\bf\sf \implies \quad \sqrt{16}\quad=\quad 4\]

Answer 53

Square roots too confusing? :o

Answer 54

no. well yeah kinda. I am just trying to type in my responses to the first question and another. so the square root of 16 is 4...

Answer 55

Yes.
So our expression further simplifies.\[\Large\bf\sf x\quad=\quad \frac{2\pm\sqrt{16}\cdot \mathcal i}{2}\qquad\to\qquad x\quad=\quad \frac{2\pm 4 \mathcal i}{2}\]Which can be further simplified.
Break it into two separate fractions,\[\Large\bf\sf x\quad=\quad \frac{2}{2}\pm\frac{4\mathcal i}{2}\]

Answer 56

Do you understand how to further simplify the fractions?

Answer 57

well if somebody asked me to show them how I would say I don't know but I am guessing the square root 16*I simplfys down to 4i and so on.

Answer 58

ya still there?

Answer 59

zepdrix???? I still need help!!!