Question About Polynomials

[Guest]

The equation Y=X^2 can be written as X^3 + PX^2 + Q = 0. Show that P=-1 and Q is equal to 20.

please help me solve this !

[HallsofIvy]

Surely this isn't the whole question? If "X^2= Y" then X^3= XY so that X^3- YX+ 0= 0. The coefficient of X^2 and the constant term are 0.

If "X^3+ PX^2+ Q= 0" with P= -1 and Q= 20, so X^3- X^2+ 20= 0. If X= -2, X^3- X^2+ 20= -8- 4+ 20= 12. If X= -3, X^3- X^2+ 20= -27- 9+ 20= -16 so there is a root between -3 and -2. The other two roots are not real numbers. What does that have to do with "Y"?

[GeradHum]

You need to rewrite Y=X2Y = X^2Y=X2 in the form X3+PX2+Q=0X^3 + P X^2 + Q = 0X3+PX2+Q=0.
If you set Y=20Y = 20Y=20, then:
X2=20⇒X3−X2+20=0X^2 = 20 \Rightarrow X^3 - X^2 + 20 = 0X2=20⇒X3−X2+20=0 fits the form with P=−1P = -1P=−1 and Q=20Q = 20Q=20.
So, P=−1P = -1P=−1 and Q=20Q = 20Q=20 as required.

[JohnGeorge]

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