Question

f(x)=2x^4-19x^2+24

Answer 1

find all zeros?

Answer 2

make factors

Answer 3

\[2x^4-19x^2+24=2x^4-16x^2-3x^2+24\]
=?

Answer 4

of 24?

Answer 5

1,6,8,12

Answer 6

shouldnt it be something like p/q

Answer 7

ok i gto possible roots 1, 1/2,2,3,3/2,4,4/2,6,6/2,8,8/2,12,12/2 but thats going to take a while, is there an easier way to solve these because i have many question similar

Answer 8

\[2 \times 24=48,3+16=19,3 \times 16=48\]

Answer 9

\[=2x^2(x^2-8)-3(x^2-8)\]
\[=(x^2-8)(2x^2-3)\]
\[=(x+\sqrt{8})(x-\sqrt{8})(\sqrt{2}x+\sqrt{3})(\sqrt{2}x-\sqrt{3})\]
\[x=-2\sqrt{2},x=2\sqrt{2},x=-\sqrt{\frac{ 3 }{ 2 }},x=\sqrt{\frac{ 3 }{ 2 }}\]

Answer 10

you reduced the equation?

Answer 11

where did you get 2*24 and the rest of those?

Answer 12

also how does 2x4−19x2+24=2x4−16x2−3x2+24