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Using factor theorem, factorize each of the following polynomials:$x^4 + 10x^3 + 35x^2 + 50x + 24$
Given:
Given expression is $x^4 + 10x^3 + 35x^2 + 50x + 24$.
To do:
We have to factorize the given polynomial.
Solution:
Let $f(x)=x^{4}+10 x^{3}+35 x^{2}+50 x+24$
The factors of the constant term 24 are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12$ and $\pm 24$
Let $x=-1$, this implies,
$f(-1)=(-1)^{4}+10(-1)^{3}+35(-1)^{2}+50(-1)+24$
$=1-10+35-50+24$
$=60-60$
$=0$
Therefore, $x+1$ is a factor of $f(x)$
Let $x=-2$, this implies,
$f(-2)=(-2)^{4}+10(-2)^{3}+35(-2)^{2}+50(-2)+24$
$=16-80+140-100+24$
$=180-180$
$=0$
Therefore, $x+2$ is a factor of $f(x)$.
Let $x=2$, this implies,
$f(2)=(2)^{4}+10(2)^{3}+35(2)^{2}+50(2)+24$
$=16+80+140+100+24$
$=360 \
eq 0$
Therefore, $x-2$ is not a factor of $f(x)$
Let $x=-3$, this implies,
$f(-3)=(-3)^{4}+10(-3)^{3}+35(-3)^{2}+50(-3)+24$
$=81-270+315-150+24$
$=420-420$
$=0$
Therefore, $x+3$ is a factor of $f(x)$
Let $x=-4$, this implies,
$f(-4)=(-4)^{4}+10(-4)^{3}+35(-4)^{2}+50(-4)+24$
$=256-640+560-200+24$
$=840-840$
$=0$
Therefore, $x+4$ is a factor of $f(x)$
Hence, $f(x)=(x+1)(x+2)(x+3)(x+4)$.