How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given $f(x)=x^4-5x^2+4$ ?

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Answer 1 · 2017-03-05T18:06:04

$(1) : f(x)" has two +ve zeroes, namely, "2 and 1;$

$(2) :" two "-ve "zeroes, "-2, and, -1:$

$(3) :" four rational zeroes, "+-2, +-1$ .

Let $x^2=y$ in $f(x)=x^4-5x^2+4=y^2-5y+4.$

$:. f(x)=0 rArr y^2-5y+4=0.$

$:. ul(y^2-4y)-ul(y+4)=0.$

$:. y(y-4)-1(y-4)=0.$

$:. (y-4)(y-1)=0.$

$:. (x^2-4)(x^2-1)=0, ...........[because, y=x^2]$

$:. (x+2)(x-2)(x+1)(x-1)=0.$

$:. x=-2, 2, -1, and, 1.$

Thus, $(1) : f(x)" has two +ve zeroes, namely, "2 and 1;$

$(2) :" two "-ve "zeroes, "-2, and, -1:$

$(3) :" four rational zeroes, "+-2, +-1$ .

Enjoy Maths.!

How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given f(x)=x^4-5x^2+4f(x)=x^4-5x^2+4?