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Bess [88]
3 years ago
9

I’ve been stuck can someone help

Mathematics
1 answer:
never [62]3 years ago
5 0

Answer:

(a). A_{x} = \left[\begin{array}{cc}12&-4\\66&6\end{array}\right]

Step-by-step explanation:

\left \{ {{5x-4y=12} \atop {3x+6y=66}} \right.

A = \left[\begin{array}{cc}5&-4\\3&6\end{array}\right]

A_{x} = \left[\begin{array}{cc}12&-4\\66&6\end{array}\right]

A_{y} = \left[\begin{array}{cc}5&12\\3&66\end{array}\right]

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On a test, Leo is asked to completely factor the polynomial 3x3 – 3x + 5x2 – 5. He uses double grouping to get (x2 – 1)(3x + 5).
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Answer:

Leo does not  factored the polynomial completely.

3x^3-3x+5x^2-5=(x+1)(x-1)(3x+5)

Step-by-step explanation:

Given: On a test, Leo is asked to completely factor the polynomial

3x^3-3x+5x^2-5.

also, he uses double grouping to get (x^2- 1)(3x+5)

We have to explain whether he  factored the polynomial completely or not.

Consider the given polynomial 3x^3-3x+5x^2-5.

after double grouping he get (x^2- 1)(3x+5)

Also, we can simplify further ,

Using algebraic identity a^2-b^2=(a+b)(a-b)

We get,

(x^2- 1)=(x+1)(x-1)

We get, (x^2- 1)(3x+5)=(x+1)(x-1)(3x+5)

Thus, 3x^3-3x+5x^2-5=(x+1)(x-1)(3x+5)

Thus, He does not  factored the polynomial completely.

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3 years ago
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Hope this helps!

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