# Find the range of all values of P.

#### Sandeep Kumar

Find the range of all possible values of B if the graph of
P(x) = 12x⁴ - 5x³ -38x² + Bx + 6
crosses the x-axis between 0 and -1

[I have concluded some points from the problem. I still need help finding the range . My working is in the attachments]

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#### romsek

MHF Helper
• topsquark

#### Idea

without using calculus?

is $$\displaystyle B$$ an integer?

#### Sandeep Kumar

without using calculus?

is $$\displaystyle B$$ an integer?
Yes, B is an integer.

#### Idea

If $$\displaystyle p(x)=12x^{4 }-5x^3-38x^2+b x+6$$ has a root $$\displaystyle r$$, $$\displaystyle -1\leq r <0$$ then $$\displaystyle b \geq -15$$

To see this , write $$\displaystyle p(r)=0$$ and show that

$$\displaystyle b+15=-\frac{(1+r) }{r}g(r)$$

where $$\displaystyle g(x)=6-21 x-17 x^2+12 x^3$$

1) $$\displaystyle g(x)$$ has a unique root $$\displaystyle \alpha$$ where $$\displaystyle -1<\alpha < -\frac{2}{3}$$

2) If $$\displaystyle r>\alpha$$ we are done. Otherwise

$$\displaystyle -2<g(r)<0$$ and $$\displaystyle 0<-\frac{(1+r) }{r}<\frac{1}{2}$$

From this it follows that $$\displaystyle b>-16$$ and again we get $$\displaystyle b \geq -15$$ since $$\displaystyle b$$ is an integer

Conversely, show that if $$\displaystyle b \geq -15$$ then

$$\displaystyle p(x)=12x^{4 }-5x^3-38x^2+b x+6$$ has a root $$\displaystyle r$$, $$\displaystyle -1\leq r <0$$

• topsquark