Recent content by RChristenk

You are right. It cannot be zero because ##b-c## is positive, ##q(b)## is negative and ##r## is negative. ##f(b)## perpetually is less than zero. So this proves if all the coefficients are negative ##b## is the upper bound? Then why does the theorem states all coefficients need to be...

According to the theorem as written, any ##f(x)## with all-negative coefficients have no bound (because the upper bound requires all non-negative signs and the lower bound requires alternating signs). But this is incorrect? Can someone explain to me please? Regarding the lower bound, let...

If all the coefficients are negative, after applying the upper bound theorem I get ##f(b)<0##. But I just realized the upper bound theorem requires the final line in the synthetic division to be strictly non-negative. So doesn't that make ##f(b)<0## meaningless?

Ah so the upper bound assumes ##f(x)## starts with all non-negative coefficients, and the lower bound assumes ##f(x)## starts with alternating signs. I kept thinking ##f(x)## was strictly positive for both cases. So if ##f(x)## alternate signs, then ##f(-x)## is positive for odd degrees and...

I understand ##f(-x)## leads to a polynomial with alternating signs. But if I use the upper bound theory on ##f(-x)##, isn't there a contradiction? The upper bound theory says the coefficients of the quotient and remainder are all non-negative for a ##c>0## divided synthetically into ##f(x)##...

To prove the upper bound: Let ##c>0##, divide it into ##f(x)## and the coefficients in the final line of the synthetic division tableau are all non-negative. Thus ##f(x)=(x-c)q(x)+r##, where ##r \geq 0## (since the coefficients are given as all non-negative) and is a constant because it's degree...

I'm thinking if a polynomial is a person, then when that person puts on clothes (the absolute value), that person is still a person. Hence I couldn't understand why the text was saying absolute values can't be written as a combination of ##x##. But you have explained clearly that by definition...

I see. But what does it mean when the text says the absolute value "is not a combination of the powers of ##x##"? Isn't ##|x^2+x+1|## for example a combination of the powers of ##x##?

According to this textbook: https://www.stitz-zeager.com/szprecalculus07042013.pdf On Page ##236## it says: But I can certainly write a combination of powers inside the absolute value: ##|x+1|, |x^2+x+1|##...etc. In fact I can put the definition of a polynomial inside the absolute value...

By definition, ##\sqrt{x^2}= \left| x \right|##. For positive ##x##, such as ##4##, it is quite straightforward: ##\sqrt{4^2}=\sqrt{16}=4##. For negative values, I am more confused: ##\sqrt{(-4)^2}=\sqrt{16}=4##. The answer will always be positive, even if you put in a negative value. So why...

Yes I think the solution is a typo too. I can see the equation is symmetric about the origin, but what do you mean the term ##-4ac## breaks the symmetry in the root? Thanks.

##x^2+ax+y^2+by=-c## ##\Leftrightarrow (x+\dfrac{a}{2})^2+(y+\dfrac{b}{2})^2=-c+\dfrac{a^2}{4}+\dfrac{b^2}{4}## The conditions under which the coefficients of this equation makes a circle: ##-c+\dfrac{a^2+b^2}{4}>0## ##\Leftrightarrow 4c < a^2+b^2## Center of circle: ##(-\dfrac{a}{2}...

Thanks for your help.

Sat-P I followed what you did. ##x^{-\frac{1}{2}}(1-2x+x^{\frac{7}{6}})=0## Set ##x^{\frac{1}{6}}=t, x=t^6## ##t^{-3}(1-2t^6+t^7)=0 \tag{1}## ##t^{-3}(1-t^6-t^6+t^6 \cdot t)=0 \tag{2}## ##t^{-3}(-t^6+1+t^6 \cdot t - t^6)=0 \tag{3}## ##t^{-3}(-(t^6-1)+t^6(t-1))=0 \tag{4}##...

The answer is 1, but it would be a strange problem indeed if it cannot be factored and is to be solved by prudent observation.