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, $f(x)=(x+1)(x+2)(x+3)(x+4)$.

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Updated on: 10-Oct-2022

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