Show that every real polynomial is equal to a product of real linear and quadratic polynomials.

Show that every real polynomial is equal to a product of real linear and quadratic polynomials.
Thanks for your help

Proof by induction (on the degree of the polynomial p(x)):
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This is clearly true for degree 1.

Now, assume that this claim is true for deg p(x) = 1, 2, ..., k.
Given a polynomial of degree k+1, we have two scenarios:

(i) k+1 is odd.
So, we have lim(x→ -∞) p(x) = -∞ and lim(x→∞) p(x) = ∞
or lim(x→ -∞) p(x) = ∞ and lim(x→∞) p(x) = -∞.

Either way, since p(x) is continuous for all x in R, we conclude that p(c) = 0 for some c in R.
Hence, p(x) = (x - c) q(x) by the Factor Theorem.

Since deg q(x) = (k+1) - 1 = k, we conclude by the (strong) inductive hypothesis that q(x) is equal to a product of linear and quadratic polynomials.

Hence, p(x) is also equal to a product of linear and quadratic polynomials.
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(ii) k+1 is even.

If p(x) has a real zero, then we are done (as in (i)).

Suppose otherwise. Then, p(x) has nonreal complex roots, which occurs in complex conjugate pairs. That is if x = a + bi is a zero, then so is x = a - bi.

By the Factor Theorem, both (x - (a + bi)) and (x - (a - bi)) are factors of p(x).
==> (x - (a + bi)) * (x - (a - bi)) = (x^2 - 2ax + (a^2 + b^2)) is a real factor of p(x).

So, p(x) = (x^2 - 2ax + (a^2 + b^2)) * q(x) for some real polynomial q(x).
Since deg q(x) = (k+1) - 2 = k - 1, we conclude by the (strong) inductive hypothesis that q(x) is equal to a product of linear and quadratic polynomials.

Hence, p(x) is also equal to a product of linear and quadratic polynomials.

This completes the induction.
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I hope this helps!