How do you find all the zeros of #f(x) = x⁴ - 10x² + 24#?

2 Answers
Jul 20, 2018

Answer:

#x = +-2" "# and #" "x = +-sqrt(6)#

Explanation:

Given:

#f(x) = x^4-10x^2+24#

Note that this quartic contains only terms of even degree, so we can start to factor it as a quadratic in #x^2#.

Note also that #10 = 4 + 6# and #24 = 4 * 6#

Hence we find:

#x^4-10x^2+24 = (x^2-4)(x^2-6)#

#color(white)(x^4-10x^2+24) = (x^2-2^2)(x^2-(sqrt(6))^2)#

#color(white)(x^4-10x^2+24) = (x-2)(x+2)(x-sqrt(6))(x+sqrt(6))#

Hence zeros:

#x = +-2" "# and #" "x = +-sqrt(6)#

graph{x^4-10x^2+24 [-5.067, 4.933, -6, 32]}

Answer:

#x=\pm2, \pm\sqrt6#

Explanation:

Given function:

#f(x)=x^4-10x^2+24#

The zeroes of above bi-quadratic polynomial is given by setting #f(x)=0# as follows

#x^4-10x^2+24=0#

#x^4-6x^2-4x^2+24=0#

#x^2(x^2-6)-4(x^2-6)=0#

#(x^2-6)(x^2-4)=0#

#x^2-6=0\ \ or\ \ x^2-4=0#

#x^2=6\ \ or\ \ x^2=4#

#x=\pm\sqrt6\ \ or\ \ x=\pm 2#

#x=\pm2, \pm\sqrt6#