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3 years ago

Multiply the polynomials (3x^2+4x+4)(2x-4)

Mathematics

2 answers:

nydimaria [60]3 years ago

7 0

Answer:

The resultant expression is $6 x^3 - 4 x^2 - 8 x - 16$ .

Step-by-step explanation:

The given expression is

$(3x^2+4x+4)(2x-4)$

Using distributive property, the above expression can be written as

$3x^2(2x-4)+4x(2x-4)+4(2x-4)$

$6x^3-12x^2+8x^2-16x+8x-16$

$6 x^3 - 4 x^2 - 8 x - 16$

Therefore the resultant expression is $6 x^3 - 4 x^2 - 8 x - 16$ .

mixas84 [53]3 years ago

5 0

Answer:

6x3−4x2−8x−16 is the answer

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Roman55 [17]

Answer:

7, 4ft

Step-by-step explanation:

Given data

area = 108 square feet

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width = (x-4)

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open bracket

$108= x^2-4x-x+4\\\\108=x^2-5x+4\\\\x^2-5x-104=0\\\\$

a= 1

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3 years ago

Carlos and Sophie, factored the trinomial –4x2 − 10x + 6. Sophie factored it as –2(2x − 1)(x + 3) and Carlos factored it as (x +

nydimaria [60]

Sophie factored it completely while Carlos did not.

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4 0

3 years ago

Write a recursive rule for the sequence. 7, 6, 13, 19, 32, ...

allsm [11]

Answer:

recursive rule for the given sequence:

$a_n = a_{n-1}+a_{n-2}$ for n > 2

Step-by-step explanation:

Given the sequence:

7, 6, 13, 19, 32, ......

then;

First term $(a_{1})$ = 7

Second term $(a_{2})$ =6

third term $(a_{3})$ = 13 and so on..

You can see that:

$a_3 = a_1+a_2 = 6+7 = 13$

similarly for:

$a_4 = a_2+a_3 = 6+13 = 19$ and so on..

The recursive rule for this sequence is:

$a_n = a_{n-1}+a_{n-2}$ for n > 2

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4 years ago

I need help on this, I have no idea what I’m doing

Studentka2010 [4]

which problems? i can teach you how to do them if you want me too.

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3 years ago