Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that every polynomial equation f(x) = 0 has at least one root, real or imaginary(complex).
Thus, $x^{6} - 3x^{4} + 2x^{2}$ + 2 = 0 has at least one root.
But $\sqrt{x}$ + 5 = 0 has no root as the given equation is not a polynomial equation, so fundamental theorem of algebra does not apply on this equation.
Note : Every polynomial equation f(x) = 0 of degree 'n' has exactly 'n' real or imaginary roots.
A root is where the polynomial equal to zero and when you graph such polynomial then the roots are the points where graph cut the x-axis. On that point y-coordinate is zero.

From the graph you can see that the polynomial has 3 roots.

Examples on fundamental theorem of algebra

(i) f(x) = -2 is a polynomial of degree 0 . This is a constant polynomial and it has zero real roots.
(ii) f(x) = x + 4 is a polynomial of degree 1 and it has one real root,
x = -4.
(iii) f(x) = $x^{2}$ + 10x + 25 = $(x + 5)^{2}$ is a polynomial of degree 2 and it has one real root, x = -5.
(iv) f(x) = $x^{3}$ + x = x($x^{2}$ + 1) is a polynomial of degree 3 and it has one real root x = 0 and two imaginary roots, x = −i, +i.

1) Factorize f(x) = $x^{4}$ – 4 using fundamental theorem of algebra.
Solution: Given f(x) = $x^{4}$ – 4
Here, n = 4, so we have 4 complex roots.
f (x) = $x^{4}$ - 4
($x^{2}$ - 2 ) ($x^{2}$ + 2)
$x^{2}$ - 2 = 0         $x^{2}$ + 2 = 0
$x^{2}$ = 2           $x^{2}$ = -2
x= $ \pm\sqrt{2}$          x = $ \pm\sqrt{2}i$
We have a factored the polynomial into four linear factors, and the zeros of f are $\sqrt{2}, -\sqrt{2},\sqrt{2}i, -\sqrt{2}i$

2) Factorize f(x) = $x^{3}$ – x using fundamental theorem of algebra.
Solution: Given f(x) = $x^{3}$ – x
f(x) = x ($x^{2}$ – 1)
Here, n = 1, so we have zero complex roots.
f(x) = x ($x^{2}$ – 1)
x = 0 and ($x^{2}$ - 1 ) = 0
$x^{2}$ - 1 = 0
$x^{2}$ = 1
x= $ \pm\sqrt{1}$
x = 1 and x = - 1
We have a factored the polynomial into three linear factors, and the zeros of f are
0, 1 and -1


11th grade math

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