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i also need to show how i got the anwer

please show how u got the anwer

2006-09-27 09:27:07 · 6 answers · asked by MICROSOFT VISTA IS THE BEST O/S! 1 in Education & Reference Homework Help

6 answers

Multiply through.

(X^2)(x^2 -2x -5x +10)

Do you see how I got that? Have you heard of FOIL? First Outside Inside Last?

(x-5)(x-2)
Multiply the first terms to get x^2
Multiply the outside terms to get -2x
Inside: -5x
Last: 10

Add all these terms together to get (x^2 -2x -5x +10)

Now that you know this, simplify this expression (combining like terms) and multiply through by x^2.

Got it?

2006-09-27 09:30:27 · answer #1 · answered by andalucia 3 · 0 0

Let's do it one step at a time:
(x-5)(x-2)
Each element will need to be multiplied with each in the other bracket:
That gives us:
x(x-2) and -5(x-2)
.
x(x-2) = x2 - 2x -------------------------A
and
-5(x-2) = -5x+10 -------------------------B
.
Hence (x-5)(x-2) is the sum of A + B:
x2 -7x + 10
.
Wait, we are only half-way through.
Your question was x2(x-5)(x-2)
We already know (x-5)(x-2) to be x2-7x+10
Hence the answer is x2(x2-7x+10)
=x4 - 7x3 + 10x2

2006-09-27 16:38:29 · answer #2 · answered by Calculus 5 · 0 0

assuming that (x2) means x squared (x^2)

multiply the two binomials
(x-5)(x-2) = x^2 - 2x-5x+10 = x^2-7x+10

now multiply that answer by x^2
x^4 - 7x^3 + 10x^2


hope that helps

2006-09-27 16:35:17 · answer #3 · answered by Deana G 5 · 0 1

Is the first part "2x" or "x squared".
For future reference, if it is x squared, you may
want to write it x^2 or sq(x).

And I agree with the first answer, just multiply.
But remember to multiply through the binomials.

2006-09-27 16:35:32 · answer #4 · answered by captn_carrot 5 · 1 0

is that x squared? x 2 times x times x is x to the 4th
-5x2 times -2x is 10x to the 3rd, so it is x to the 4th plus 10x to the 3rd.

2006-09-27 16:34:08 · answer #5 · answered by Steve C 3 · 0 1

(x2)(x-5)(x-2) =

(x2) [(x-5)(x-2)] =

(x2) [x2-2x-5x+10] =

(x2) (x2-7x+10) =

x4-7x3+10x2

2006-09-27 16:43:13 · answer #6 · answered by Ms. Francoise 2 · 0 1

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