"Given P(x)=3x5−4x4+9x3−12x2−12x+16, and that 2i is a zero,..." solved and explained

High School
Algebra
Pre-Algebra
Factors and multiples

e1s2kat26

Answered question

2021-08-06

Given

P (x) = 3 x^{5} - 4 x^{4} + 9 x^{3} - 12 x^{2} - 12 x + 16

, and that 2i is a zero, write P in factored form (as a product of linear factors). Be sure to write the full equation, including

P (x) =

Answer & Explanation

okomgcae

Skilled2021-08-07Added 93 answers

$P (x) = 3 x^{5} - 4 x^{4} + 9 x^{3} - 12 x^{2} - 12 x + 16$
Given that,
2i is a zero of P(x)
$∴ - 2 i$ is also a zero of P(x)
$∴ (x - 2 i), (x + 2 i)$ are two factors of P(x)
Now $(x - 2 i) (x + 2 i) = x^{2} - {(2 i)}^{2}$
$= x^{2} - 4 i^{2}$
$x^{2} + 4$
$∴ P (x) = 3 x^{5} - 4 x^{4} + 9 x^{3} - 12 x^{2} - 12 x + 16$
$= 3 x^{3} (x^{2} + 4) - 4 x^{2} (x^{2} + 4) - 3 x (x^{2} + 4) + 4 (x^{2} + 4)$
$= (x^{2} + 4) (3 x^{3} - 4 x^{2} - 3 x + 4)$
$= (x^{2} + 4) [3 x^{2} (3 x - 4) - 1 (3 x - 4)]$
$= (x^{2} + 4) (3 x - 4) (x^{2} - 1)$
$= (x^{2} + 4) (3 x - 4) (x + 1) (x - 1)$
$= (x + 2 i) (x - 2 i) (3 x - 4) (x + 1) (x - 1)$
For zeros,
$P (x) = 0$
$\Rightarrow 3 x^{5} - 4 x^{4} + 9 x^{3} - 12 x^{2} - 12 x + 16 = 0$
$\Rightarrow (x + 2 i) (x - 2 i) (3 x - 4) (x + 1) (x - 1) = 0$
$∴$ Zeros of P(x) are
$- 2 i, 2 i, \frac{4}{3}, - 1, 1$ .

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