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2 years ago

30 points!!! √8+√98+√32.

Mathematics

2 answers:

frez [133]2 years ago

6 0

Let us first resolve the given numbers into prime factors.

By prime factorization,

$\longmapsto$ √8 = √( 2 × 2 × 2)

$\longmapsto$ √8 = 2√2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longmapsto$ √98 = √( 2 × 7 × 7)

$\longmapsto$ √98 = 7√2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longmapsto$ √32 = √( 2 × 2 × 2 × 2 × 2)

$\longmapsto$ √32 = 2 × 2√2

$\longmapsto$ √32 = 4√2

According to the question,

$\longmapsto$ √8+√98+√32

$\longmapsto$ 2√2 + 7√2 + 4√2

Taking √2 as common.

$\longmapsto$ √2(2 + 7 + 4)

$\longmapsto$ √2(13)

$\longmapsto$ 13√2

So, we get that :

yanalaym [24]2 years ago

5 0

Answer:

13√2 or 18.38478

Step-by-step explanation:

i hope this helps :)

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Multiplying polynomials (3m-1)(8m+7)

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3 years ago

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What is the result when x^3+ 7x^2+ 9x - 8 is divided by x+ 2?

Ahat [919]

Answer:

$\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$

So the quotient is $1x^2+5x+-1$ and the remainder is $-6$ .

Step-by-step explanation:

We could do this by synthetic division since the denominator is a linear factor in the form $x-c$ .

Since we are dividing by $x+2=x-(-2)$ , this is our setup for the synthetic division:

-2 | 1 7 9 -8

| -2 -10 2

______________

1 5 -1 -6

So the quotient is $1x^2+5x+-1$ and the remainder is $-6$ .

So $\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$ .

We can also do long division.

x^2+5x-1

____________________

x+2| x^3+7x^2+9x-8

-(x^3+2x^2)

-------------------

5x^2+9x-8

-( 5x^2+10x)

--------------------

-x-8

-(-x-2)

--------------

-6

So we see here we get the same quotient, $x^2+5x-1$ . and the same remainder, $-6$ .

Now let's check our result that:

$\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$ .

So I'm going to rewrite the right hand side as a single fraction:

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)}{x+2}+\frac{-6}{x+2}$ .

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}$

Now let's focus on multiplying $(x+2)(x^2+5x-1)$ .

We are going to multiply the first term of the first ( ) to every term in the second ( ).

We are also going to multiply the second term of the first ( ) to every term in the second ( ).

$x(x^2)=x^3$

$x(5x)=5x^2$

$x(-1)=-x$

$2(x^2)=2x^2$

$2(5x)=10x$

$2(-1)=-2$

---------------------------Combine like terms:

$x^3+(5x^2+2x^2)+(-x+10x)+-2$

$x^3+7x^2+9x-2$

So let's go back where we were in our check of $\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$ :

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}$

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-2-6}{x+2}$

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-8}{x+2}$

We have the exact same thing on both sides so we did good.

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Step-by-step explanation:

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30 points!!! √8+√98+√32.​

30 points!!! √8+√98+√32.