Explanation:
that feels really so difficult to you ?
what is it you don't understand ? because this is really very easy. we take the input value as x and simply calculate the result.
1
f(x) = x³ - 5x² + 6x - 4
f(2) = 2³ - 5×2² + 6×2 - 4 = 8 - 20 + 12 - 4 = -4
2
f(x) = 4x³ + 3x² + x + 2
f(1) = 4×1³ + 3×1² + 1 + 2 = 4 + 3 + 1 + 2 = 10
3
f(x) = 2x⁴ - x³ + 3x² - 1
f(-1) = 2×(-1)⁴ - (-1)³ + 3×(-1)² - 1 = 2×1 + 1 + 3 - 1 = 5
4
f(x) = 2x³ - 6x - 5
f(-3) = 2×(-3)³ - 6×-3 - 5 = -54 + 18 - 5 = -41
5
f(x) = x³ - 4x² - x
f(4) = 4³ - 4×4² - 4 = (4³ - 4³) - 4 = -4
6
f(x) = x³ + 2x² - 2x - 1 divided by x -1 = x² + 3x + 1
- x³ - x²
-----------------------
0 + 3x² - 2x
- 3x² - 3x
---------------------------
0 + x - 1
- x - 1
--------------------------------
0 remainder, so, yes, x-1 is a factor.
7
f(x) = 4x² + 13x + 10 divided by x + 2 = 4x + 5
- 4x² + 8x
-----------------------
0 + 5x + 10
- 5x + 10
‐------------------------------
0 remainder, so x + 2 is a factor.
8
let's now try the factor theorem (the term is a factor if the zero point of the term is also a zero point of the whole function).
the zero point of x - 2 is x = 2.
f(x) = 4x² + 13x + 10
f(2) = 4×2² + 13×2 + 10 = 16 + 26 + 10 = 52 and not 0, therefore, x - 2 is NOT a factor.
9
the zero point of x + 3 is x = -3
f(x) = 3x³ + 10x² + x - 6
f(-3) = 3×(-3)³ + 10×(-3)² + (-3) - 6 = -81 + 90 - 3 - 6 = 0
so, x + 3 is a factor.
10
the zero point of x - 5 is x = 5.
so, f(5) must be 0 for x-5 to be a factor.
f(x) = 2x³ - 13x² + kx + 10
f(5) = 2×5³ - 13×5² + 5k + 10 = 250 - 325 + 5k + 10 =
= -65 + 5k
and that has to be 0.
0 = -65 + 5k
65 = 5k
k = 13
so, x-5 is a factor of 2x³ - 13x² + 13x + 10