Question
$x^4 + 10x^3 + 35x^2 + 50x + 24.$
✓
Answer
Let $f(x)=x^4+10 x^3+35 x^2+50 x+24$
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,
$f(x)=x^3(x+1)+9 x^2(x+1)+26(x+1)+24(x+1)$
$=(x+1)\left(x^3+9 x^2+26 x+24\right)$
$=(x+1) g(x) \ldots(1)$
Where $g(x)=x^3+9 x^2+26 x+24$
Putting $x=-2$, we get:
$g(-2)=(-2)^3+9(-2)^2+26 x(-2)+24$
$=-8+36-52+24=60-60$
$=0$
Therefore, $(x+2)$ is the factor of $g(x)$.
Now,
$g(x)=x^2(x+2)+7 x(x+2)+12(x+2)$
$=(x+2)\left(x^2+7 x+12\right)$
$=(x+2)\left(x^2+4 x+3 x+12\right)$
$=(x+2)(x+3)(x+4) \ldots(2)$
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,
$f(x)=x^3(x+1)+9 x^2(x+1)+26(x+1)+24(x+1)$
$=(x+1)\left(x^3+9 x^2+26 x+24\right)$
$=(x+1) g(x) \ldots(1)$
Where $g(x)=x^3+9 x^2+26 x+24$
Putting $x=-2$, we get:
$g(-2)=(-2)^3+9(-2)^2+26 x(-2)+24$
$=-8+36-52+24=60-60$
$=0$
Therefore, $(x+2)$ is the factor of $g(x)$.
Now,
$g(x)=x^2(x+2)+7 x(x+2)+12(x+2)$
$=(x+2)\left(x^2+7 x+12\right)$
$=(x+2)\left(x^2+4 x+3 x+12\right)$
$=(x+2)(x+3)(x+4) \ldots(2)$
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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