Find all zeros of x^4+x

Mathematics

1 answer:

Yuri [45]1 year ago

4 0

Answer:

-1 and 0

Step-by-step explanation:

X ^4+x = 0;

x*(x^3+1)=0;

x*[(x^3-x)+(x+1)]=0;

x*[x(x-1)(x+1)+(x+1)]=0;

x*(x+1)(x^2-x+1)=0;

Since x^2-x+1=0 has no intersection with the X-axis, there is no x value such that y=0

x=0 or -1

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Hey! Can someone check my answers?

Andrew [12]

The greatest common factor is the biggest number taken from the values.

Q1. The answer is <span>A. 5y^6
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Since $x^{a}* x^{b} =x^{a+b}$ , then $x^{9}= x^{3}* x^{6}$

Back to our expression:
$4*5 y^{9}+5 y^{6}=4*5 y^{3}*y^{6}+5 y^{6}=4 y^{3}*5y ^{6}+ 5y ^{6}*1=5 y^{6} (4y ^{3} +1)$
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Q2. The answer is <span>D. 12xy^2
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We will use the rule $x^{a}* x^{b} =x^{a+b}$ to factorise the exponents:
$12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}= \\ =12x*y^{2}*y^{3}+5*12*x* x^{3} *y^{2}-2*12x* x^{2} *y*y^{2}= \\ =12xy^{2}*y^{3}+12xy^{2}*5 x^{3} -12xy^{2}*2 x^{2} y= \\ =12xy^{2}(y^{3}+5 x^{3}-2 x^{2} y)$
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Hey! I'v been struggling with this question lately. Can someone just lend me a hand?

Dominik [7]

So, this problem is asking us to find the value of our variable u.

How can you do this? By isolating the variable, or having only it on one side of the equals sign.

Here's the step by step approach. It's just simple algebra and reverse order of operation (which is what you should do whenever you are solving for a variable).

$\frac{u+88}{11} = 17$

Multiply both sides by 11 to get rid of the 11 denominator.
$u + 88 = 187$

Then, subtract both sides by 88 to get u by itself.
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Therefore, u is equal to 99.

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