Part A:The area of a square is (4x2 − 12x + 9) square units. Determine the length of each side of the square by factoring the area expression completely. Show your work. (5 points)
Part B: The area of a rectangle is (16x2 − 9y2) square units. Determine the dimensions of the rectangle by factoring the area expression completely. Show your work. (5 points)

- anonymous

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- anonymous

@Mackenzie_Willa

- anonymous

@welshfella

- anonymous

I have the answer just not the steps!
Part A:(2x -3)^2
Part B:(4x+3y)(4x−3y)

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## More answers

- anonymous

@Nnesha

- amilapsn

Do you know these?
\[a^2-b^2=(a-b)(a+b)
\\
(a+b)^2=a^2+2ab+b^2\]

- anonymous

Because the first shape is a square, the length and the width must be the same. So the area must be a perfect square. We need to find an expression that , when multiplied by itself, will give \(4x^2-12x+9\)Start by taking the square roots of the first and last coefficients. Can you do that?

- anonymous

Can you help me work it ut step by step plz

- anonymous

*out

- anonymous

What is the square root of the first term, \(4x^2\)?

- anonymous

2x^2?

- anonymous

Not quite. The square of 4 is 2, so you got that part right, but what is the square root of x^2? What can you multiply by itself to give x^2?

- anonymous

*square root of 4

- anonymous

I am confused

- anonymous

Do you understand that\[x \times x = x^2\]

- anonymous

yes

- anonymous

OK. So the square root of \(x^2\) is \(x\). OK with that?

- anonymous

oh okay.

- anonymous

All right. So the square root of \(4x^2\) is \(2x\). In other words\[2x \times 2x = 4x^2\]Make sense now?

- anonymous

Yes

- anonymous

OK. So now you need the square root of the last term, +9. What is it?

- anonymous

3

- anonymous

Great. With these two square roots, you have determined that the two factors must be\[4x^2-12x+9 = \left( 2x \text{ ? }3 \right)\left( 2x \text{ ? }3 \right)\]All that is left is to choose the correct sign, plus or minus. For that, look at the second term in the given trinomial. What sign does it have?

- anonymous

Minus.

- anonymous

Perfect. Then that is the sign you use in the factors. Therefore,\[4x^2-12x+9 = \left( 2x-3 \right)\left( 2x-3 \right)\]and you have your answer.

- anonymous

Keep in mind that this technique only works for factoring a perfect square trinomial.

- anonymous

Are these the answers. I have been working them out too:
Part A)
4x2 − 12x + 9
4x^2 -6x -6x + 9
2x( 2x - 3) -3(2x - 3)
(2x - 3)(2x-3)
length of each side is 2x-3
Part B)
16x^2 − 9y^2
(4x)^2 - (3y)^2
(4x - 3y) (4x+3y)
dimensions are 4x-3y, 4x+3y.

- anonymous

Yes, you are correct. Very well done!

- anonymous

Thank you!

- anonymous

You're welcome.

- anonymous

@ospreytriple can you help with 1 more?

- anonymous

An expression is shown below:
f(x) = –16x2 + 60x + 16
Part A: What are the x-intercepts of the graph of the f(x)? Show your work. (2 points)
Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points)
Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

- anonymous

Part A: To find the x-intercepts, you must solve the equation \(-16x^2+60x+16=0\). Which method would you use; factoring by completing the square, or the quadratic formula?

- anonymous

quadratic formula?

- anonymous

OK. Go ahead. What do you get?

- anonymous

Can you refresh me on the formula.

- anonymous

For the general quadratic, \(ax^2 + bx + c=0\), the x-intercepts are found using\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

- anonymous

can you help me through this one

- anonymous

I am really confused.

- anonymous

Have you used the quadratic formula before?

- anonymous

Just learning it.

- anonymous

OK. I'll help you set it up.

- anonymous

Okay thank you! I do have 1 request can you say like part a and like part b and part c because i will get confused and don't know what to say.

- anonymous

This is for Part A. If you look at the given trinomial and compare it with the general quadratic, you'll see that \(a=-16\), \(b=60\) and \(c=16\). Putting these values into the quadratic formula, you have\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{2a } = \frac{ -60 \pm \sqrt{60^2-4\left( -16 \right)\left( 16 \right)} }{ 2\left( -16 \right) }\]Can you complete these calculations?

- anonymous

is it -5?

- anonymous

One second. I'm working it out...

- anonymous

okay ;]

- anonymous

Wait is it...
x = 4 and x = -1/4.

- anonymous

Yayyyy! You got it. Those are the x-intercepts, \(x=4\) and \(x=-\frac{1}{4}\)

- anonymous

Yay :V.
can you help with part B and C too?

- anonymous

Part B: Vertex a maximum or minimum? What it means is, is the parabola opening upward or downward? If the parabola opens upward, then the vertex is a minimum. If the parabola opens downward, then the vertex is a maximum.|dw:1440091745676:dw|
Do you know how to tell which direction the parabola opens?

- anonymous

These are my answers so far are they correct?
Part A.)
f(x) = –16x^2 + 60x + 16
-4(x-4)(x+1/4)
x-4=0
x+1/4=0
x=4 and x=-1/4
Part B.)
vertex (15/8, 289/4) it is a maximum because the coeffficient of x^2 is -16 which is negative
Part C.)
You can graph the x intercepts and the vertex. That will give you the overall shape of the function.

- anonymous

OK. That saves me a lot of work. Hope I'm not wasting your time. I see in Part A you factored instead of using the quadratic formula. Parts B & C are correct. Good job>

- anonymous

Part c is correct?

- anonymous

Yes.

- anonymous

Oh. I wrote that down for notes rofl

- anonymous

Thank you!

- anonymous

You're welcome

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