Question

Let $p(z)$ be a polynomial of degree $n ge 1$ and let $z_0 in mathbb{C}$. Show that there is a polynomial $h(z)$ of degree $n - 1$ such that $p(z) = (z - z_0)h(z) + p(z_0)$. In particular, if $p(z_0) = 0$, then $p(z) = (z - z_0)h(z)$.

          Let $p(z)$ be a polynomial of degree $n ge 1$ and let $z_0 in mathbb{C}$. Show that there is a polynomial $h(z)$ of degree $n - 1$ such that $p(z) = (z - z_0)h(z) + p(z_0)$. In particular, if $p(z_0) = 0$, then $p(z) = (z - z_0)h(z)$.
        
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Let p(z) be a polynomial of degree n ge 1 and let z0 in mathbbC. Show that there is a polynomial h(z) of degree n - 1 such that p(z) = (z - z0)h(z) + p(z0). In particular, if p(z0) = 0, then p(z) = (z - z0)h(z).

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Let p(z) be a polynomial of degree n ≥ 1 and let z0 ∈ ℂ. Show that there is a polynomial h(z) of degree n - 1 such that p(z) = (z - z0)h(z) + p(z0). In particular, if p(z0) = 0, then p(z) = (z - z0)h(z).
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Transcript

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00:01 Okay, so we are going to prove the claim in our exercise by induction.
00:07 So induction.
00:11 Okay, well, we are going to proceed by induction on the degree of our polynomial p of z.
00:19 Now, if the degree of our polynomial p of z is equal to one, then this thing implies what? this thing implies that p of z is a linear polynomial so in particular this thing shows that p of z is going to be of the form a z plus b with a different from zero now how can we write a z plus b well this is actually pretty easy we can write z as okay we can write z as z as z minus z 0 plus z zero so p of z is going to be equal to what well this guy is going to be equal to a multiplied by z minus z plus okay so here we're going to have plus a multiplied by z 0 plus b perfect so our claim is true so claim through for degree of p of z equals 1 perfect now let's show that our claim true for n minus 1 implies our claim true true for n so claim true for n.
02:09 Okay, perfect.
02:11 Now we have a polynomial p of z of degree n.
02:17 Now this one is a complex polynomial, that is a polynomial with complex coefficients.
02:23 So in particular, we can find z star, which is going to be a root of p of z.
02:38 Because we know that every polynomial with complex coefficients has at least a root.
02:45 Actually, we can always factorize this polynomial as a product of linear polynomials because we are working on the field of complex numbers.
02:56 Perfect.
02:57 That being said, p of z can be written as what? it can be written as q of z multiplied by z minus z star.
03:08 Where the degree of our polynomial q is n minus 1.
03:14 Perfect...
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