Solve for p.

Mathematics

2 answers:

FromTheMoon [43]4 years ago

6 0

Answer:

Option C.

Step-by-step explanation:

The given inequality is

$2p+3>2(p-3)$

We need to find the solutions of this inequality.

Using distributive property we get

$2p+3>2(p)+2(-3)$

$2p+3>2p-6$

Subtract 2p from both sides.

$2p+3-2p>2p-6-2p$

$3>-6$

This inequality is free from variable p and it is true for any value of p.

The solution set of given inequality is the set of all real numbers. Therefore the correct option is C.

Nina [5.8K]4 years ago

5 0

All real numbers............

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Notice that when $x=-1$ (which you can arrive at via the rational root theorem), you have

$(-1)^5+3(-1)^4+(-1)^3-5(-1)^2-6(-1)-2=-1+3-1-5+6-2=0$

which means you can pull out a factor of $x+1$ . Upon dividing you get

$\dfrac{x^5+3x^4+x^3-5x^2-6x-2}{x+1}=x^4+2x^3-x^2-4x-2$

The rational root theorem will come in handy again, suggesting that $x=-1$ appears a second time as a root, which means

$\dfrac{x^4+2x^3-x^2-4x-2}{x+1}=x^3+x^2-2x-2$

Now this is more readily factored without having to resort to the rational root theorem. You have

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

so in fact, $x=-1$ shows up as a root for a third time.

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Two roots are obvious, $x=0$ and $x=-1$ (with multiplicity 3). The remaining two are given by

$x^2-2=0\implies x^2=2\implies x=\pm\sqrt2$