Question
$x^4 + 10x^3 + 35x^2 + 50x + 24.$
✓
Answer
Let $f(x) = x^4 + 10x^3 + 35x^2 + 50x + 24$
Now, putting $x = -1,$
we get $f(-1) = (-1)^4 + 10(-1)^3 + 35(-1)^2 + 50(-1) + 24$
$= 1 - 10 + 35 - 50 + 24$
$= 60 - 60$
$= 0$
Therefore, $(x + 1)$ is a factor of polynomial $f(x).$
Now, $f(x) = x^3(x + 1) + 9x^2(x + 1) + 26(x + 1) + 24(x + 1)$
$= (x + 1)(x^3 + 9x^2 + 26x + 24)$
$= (x + 1)g(x) ...(1)$
Where $g(x) = x^3 + 9x^2 + 26x + 24$
Putting $x = -2,$
we get: $g(-2)$
$= (-2)^3 + 9(-2)^2 + 26x(-2) + 24$
$= -8 + 36 - 52 + 24$
$= 60 - 60$
$= 0$
Therefore, $(x + 2)$ is the factor of $g(x).$
Now, $g(x) = x^2(x + 2) + 7x(x + 2) + 12(x + 2)$
$= (x + 2)(x^2 + 7x + 12)$
$= (x + 2)(x^2 + 4x + 3x + 12)$
$= (x + 2)(x + 3)(x + 4) ...(2)$
From equation $(1)$ and $(2),$
we get:$ f(x) = (x + 1)(x + 2)(x + 3)(x + 4)$
Hence, $(x + 1), (x + 2), (x + 3)$ and $(x + 4)$ are the factors of polynomial $f(x).$
Now, putting $x = -1,$
we get $f(-1) = (-1)^4 + 10(-1)^3 + 35(-1)^2 + 50(-1) + 24$
$= 1 - 10 + 35 - 50 + 24$
$= 60 - 60$
$= 0$
Therefore, $(x + 1)$ is a factor of polynomial $f(x).$
Now, $f(x) = x^3(x + 1) + 9x^2(x + 1) + 26(x + 1) + 24(x + 1)$
$= (x + 1)(x^3 + 9x^2 + 26x + 24)$
$= (x + 1)g(x) ...(1)$
Where $g(x) = x^3 + 9x^2 + 26x + 24$
Putting $x = -2,$
we get: $g(-2)$
$= (-2)^3 + 9(-2)^2 + 26x(-2) + 24$
$= -8 + 36 - 52 + 24$
$= 60 - 60$
$= 0$
Therefore, $(x + 2)$ is the factor of $g(x).$
Now, $g(x) = x^2(x + 2) + 7x(x + 2) + 12(x + 2)$
$= (x + 2)(x^2 + 7x + 12)$
$= (x + 2)(x^2 + 4x + 3x + 12)$
$= (x + 2)(x + 3)(x + 4) ...(2)$
From equation $(1)$ and $(2),$
we get:$ f(x) = (x + 1)(x + 2)(x + 3)(x + 4)$
Hence, $(x + 1), (x + 2), (x + 3)$ and $(x + 4)$ are the factors of polynomial $f(x).$
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