Assuming that h(x) = 7x4 - 9x3 - 41x2 + 13x + 6 and h(-2) = 7, (-2) 4 - 9(-2) (-2) 3 - 41(-2) (-2) 2 + 13(-2) + 6 = 7(16) - 9(-8) (-8) - 41(4) + 13(-2) + 6 = 0;
This indicates that (x + 2) is a synthetic polynomial factor.
(7x4 - 9x3 - 41x2 + 13x + 6) ÷ (x + 2):
The balance is zero.
A polynomial of degree 3 makes up the quotient: g(x) = 7x3 - 23x2 + 5x + 3.
The Rational Zeros Theorem shows that p: 1, 3, which are factors of the constant term 3, and q: 1, 7, which are factors of the leading coefficient 7, respectively
All potential outcomes are p/q: 1, 3, 1/ 7, and 3/7 g(3) = 7. (3)
3 - 23(3) +(3)
By artificial division:
The balance is zero.
The polynomial quotient is quadratic.
7x2 - 2x - 1 = 0
Equation for quadratics = [-b (b2 - 4ac)]/2a
x = -(-2) ± √[(-2)2 - (4 × 7 × -1)]/ (2 × 7) = (2 ± √32)/14 = (1 ± 2√2)/7
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