What is the wrong with the problem
1 answer:
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Sum of 2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3
Answer:
• zero: -4, -4/3, 7
• positive: -4 < x < -4/3 . . . or 7 < x
• negative: x < -4 . . . or -4/3 < x < 7
Step-by-step explanation:
Zeros of the function are at x=-4, -4/3, +7. These are the values that make each of the individual factors be zero. For example, x-7=0 when x=7.
The function will be negative for x-values left of an odd number of zeros. It will be positive for x-values left of an even number of zeros (including left of no zeros, which is to say right of all zeros). This is because the sign of the factor giving rise to the zero changes for x-values on either side of that zero. (This is not true for zeros with even multiplicity, as the sign does not change at those.)
