Answer:
x³ - 6·x² + 6·x + 8 = (x - 4) × (x - (1 + √3)) × (x + 1 + √3)
Step-by-step explanation:
From the given graph of the polynomial, x³ - 6·x² + 6·x + 8, we have that one of the zeros of the polynomial, which is one point the graph cuts the x-axis is given as follows;
At a zero, we have, x = 4
∴ x - 4 = 0 is a factor of the polynomial
Dividing the given polynomial by (x - 4), gives;
<u>x² - 2·x - 2</u>
(x³ - 6·x² + 6·x + 8)÷(x - 4)
<u>x³ -4·x²</u>
-2·x² + 6·x + 8
<u>-2·x² + 8·x</u>
-2·x + 8
<u>-2·x + 8</u>
<u>0</u>
∴ (x³ - 6·x² + 6·x + 8)÷(x - 4) = x² - 2·x - 2
∴ (x - 4) × (x² - 2·x - 2) = x³ - 6·x² + 6·x + 8
Factorizing x² - 2·x - 2, gives;
x = (2 ± √((-2)² - 4×1×(-2)))/(2 × 1) = (2 ± 2·√(3)/(2) = (1 ± √(3)
Therefore, x = 1 + √3, (x - (1 + √3)) = 0 or x = 1 - √3, (x - (1 - √3)) = 0 which gives the factors, after factorizing completely as follows;
x³ - 6·x² + 6·x + 8 = (x - 4) × (x - (1 + √3)) × (x + 1 + √3)