3(x -y + z)(x + y +z)?

Question

Here am I supposed to multiply both equations by 3?

Because If so I get 3x -3y +3z
3x +3y +3z
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I get stuck here. The answer is 3x^2 +6xz -3y^2 +z^2. Can someone pls explain the steps? Thanks.

Answer 1

3(4)(5) = 12(5) = 3(20) = 60

no, only multiply 1 of them by 3. this is associative property of multiplication, not distributive property.

I'd multiply out the polynomials, then multiply by 3:

3(x - y + z)(x + y + z) =
3(x² + xy + xz - xy - y² - yz + xz + yz + z²) =
3(x² - y² + z² + 2xz) =
3x² - 3y² + 3z² + 6xz

you forgot to type the 3 in front of z²

Answer 2

For the question 3(x - y + z)(x + y + z), here's what you do:

First, you use the distributive property in the parenthesis: We will begin with the variable x in the first parenthesis:

x * x = x^2
x * y = xy
x * z = xz

Next, we multiply -y:

-y * x = -xy
-y * y = -y^2
-y * z = -yz

Then, we multiply z:

z * x = xz
z * y = yz
z * z = z^2

Now, you organize them all together:

x^2 + xy + xz + -xy - y^2 - yz + xz + yz + z^2

Then, we will combine like terms:

x^2 - y^2 + z^2 + xy - xy + xz + xz - yz +yz

We are now left with the following:

x^2 - y^2 + z^2 +2xz

We cancelled out xy and yz.

Now, we have 3 (x^2 - y^2 + z^2 + 2xz)

We then multiply each one by 3

3 * x^2 = 3x^2
3 * y^2 = -3y^2
3 * z^2 = 3z^2
3 * 2xz = 6xz

Now we have the answer: 3x^2 + 6xz - 3y^2 + z^2

That's all the steps required to derive the answer. Good luck! :-)

Answer 3

You have to multiply out the brackets first!

Step 1 - Multiply out the brackets:

3 ( x - y + z )( x + y + z )
= 3 ( x^2 + xy +xz - yx - y^2 - yz + zx +zy + z^2 )

Step 2 - Tidy up the numbers in the bracket:

3 ( x^2 + 2xz - y^2 + z^2 )

Step 3 - Multiply the numbers in the bracket by the 3 and that gives your answer:

3x^2 + 6xz - 3y^2 + z^2

Hope that explains it!

Answer 4

ever done foiling? multiply (x-y+z) times (x+y+z). you would get:
x^2 + xy+ xz + -y^2 + -yx + -yz + zx + zy + z^2
ever canceled things out? xy and -yx cancel, and so do -yz and zy. so taking out the canceled things means:
x^2 + xz + zx -y^2 + z^2
now here's where the 3 comes in. multiply everything by 3
3(x^2 + xz + zx + y^2 + z^2) becomes 3x^2 + 3xz + 3zx + -3y^2 + 3z^2
3xz + 3zx = 6xz (order with the letters doesnt matter. each letter matches up with each letter. that's why the other things cancelled.)
3x^2 + 6xz + -3y^2 + 3z^2

Answer 5

Multiplication is commutative, so work the brackets first.

3(x -y + z)(x + y +z)
= 3 * (x^2 +` xy + xz - yx - y^2 - yz + zx + zy + z^2)

Now bring like terms together ... (Note xy is the same as yx)...
= 3 * (x^2 +` xy - yx +xz + zx - yz + zy - y^2 + z^2)
= 3 * (x^2 + 2xz - y^2 + z^2)
= 3x^2 + 6xz - 3y^2 + 3z^2

Answer 6

Why not use the difference of two squares?

3(x -y + z)(x + y +z)
= 3[(x+z)^2-y^2]
= 3[x^2+2xz+z^2-y^2]
= 3x^2 +6xz -3y^2 +6z^2.

Answer 7

Expand & simplify the expression.

FirsT: combine the 1st two terms - use the Distribution Method.

3(x -y + z) = 3(x)+3(-y)+3(z) = = 3x-3y+3z

you have... (3x-3y+3z)(x+y+z)

Sec: combine the last two terms - use the Foil Method.

(3x)(x)+(3x)(y)+(3x)(z)+(-3y)(x)+(-3y)(y)+(-3y)(z)+(3z)(x)+(3z)(y)+(3z)(z)

= 3x^2+3xy+3xz-3xy-3y^2-3yz+3xz+3yz+3z^2

Third: combine "like" terms.

= 3x^2+3xy-3xy+3xz+3xz-3y^2-3yz+3yz+3z^2

= 3x^2 + 6xz - 3y^2 + 3z^2

and the expression has to be placed in alpha descending order...which is, 3x^2 - 3y^2 + 3x^2 + 6xz

P.S. the result is not 3x^2 +6xz -3y^2 +z^2

Answer 8

3(x-y+z)(x+y+z)
Multiply the two trinomials first
3(x^2+xy+xz-xy-y^2-yz+xz+yz+z^3)
Simplify
3(x^2+2xz-y^2+z^3)
Distribute the 3
3x^2+6xz-3y^2+3z^2

Answer 9

No, you can multiply the three by either quantity, but only 1 of the quantities.
(3x-3y+3z)(x+y+z)
3x^2+3xy+3xz-3xy-3y^2-3yz+3xz+3yz+3z^2
3x^2+3xz-3y^2+3z^2

Answer 10

You have 3 a b. First multiply a by b and then, by 3


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