Find all roots of the polynomial P(z) = z3 - 6z2 + 13z − 10 and factor it into linear factors.

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Step 1: Find a root of the polynomial P(z) = z³ − 6z² + 13z − 10 using the Rational Root Theorem. The possible rational roots are the divisors of the constant term (−10), which are ± 1, ± 2, ± 5, ± 10. Let's test z=2: P(2) = (2)³ − 6(2)² + 13(2) − 10; P(2) = 8 − 6(4) + 26 − 10; P(2) = 8 − 24 + 26 − 10; P(2) = 34 − 34 = 0; Since P(2)=0, z=2 is a root, and (z−2) is a linear factor. Step 2: Divide the polynomial P(z) by (z−2) to find the quadratic factor. We can use synthetic division: 2 | 1 -6 13 -10 | 2 -8 10 ----------------- 1 -4 5 0 The quotient is z² − 4z + 5. So, z³ − 6z² + 13z − 10 = (z−2)(z² − 4z + 5). Step 3: Factor the quadratic term z² − 4z + 5. We use the quadratic formula z = (−b ± √(b² − 4ac))/(2a) for az² + bz + c = 0. Here, a=1, b=−4, c=5. Calculate the discriminant Δ = b² − 4ac: Δ = (−4)² − 4(1)(5); Δ = 16 − 20; Δ = −4 Now, find the roots: z = (−(−4) ± √(−4))/(2(1)); z = (4 ± 2i)/(2); z = 2 ± i; The two complex roots are z₁ = 2+i and z₂ = 2−i. Therefore, the linear factors for the quadratic term are (z − (2+i)) and (z − (2−i)). Step 4: Write the complete factorization into linear factors. The linear factors are (z−2), (z−(2+i)), and (z−(2−i)). z³ − 6z² + 13z − 10 = (z−2)(z−(2+i))(z−(2−i)); z³ − 6z² + 13z − 10 = (z−2)(z−2−i)(z−2+i); The factorization of z³ − 6z² + 13z − 10 = 0 into linear factors is: (z−2)(z−2−i)(z−2+i)

Point	z	y
y-intercept	0	−10
x-intercept	2	0

Find all roots of the polynomial P(z) = z3 − 6z2 + 13z − 10 and factor it into linear factors.

Answer

Step-by-step solution

Find a root of the polynomial $P(z) = z^3 - 6z^2 + 13z - 10$ using the Rational Root Theorem.

Divide the polynomial $P(z)$ by $(z-2)$ to find the quadratic factor.

Factor the quadratic term $z^2 - 4z + 5$ .

Write the complete factorization into linear factors.

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