Question

Factorise x ki power 4 + x cube minus 7 x square minus x + 6 by factor thorm

Answer 1

Your Question:

Factorize $\tt x^4+x^3-7x^2-x+6$ by factor theorem.

Your Answer:

We have to first notice the constant term that is 6 here.

So, factors of 6 are $\tt\pm1,\pm2,\pm3,\pm6$

Now using hit and trial method.

Taking the value of x as 1

$\tt p(1)=1^4+1^3-7(1)^2-1+6 \\\\ \tt =1+1-7-1+6 \\\\ \tt =0$

So, x=1 is a zero

so, (x-1) is a factor.

Now again hit and trial method

taking value of x as 2

$\tt p(2)=2^4+2^3-7(2)^2-2+6 \\\\ \tt =16+8-28-2+6 \\\\ \tt =0$

So, x=2 is a zeros

so, (x-2) is a factor

Now again hit and trial method

taking value of x as 3

$\tt p(3)=3^4+3^3-7(3)^2-3+6 \\\\ \tt =81+27-63-3+6 \\\\ \tt \neq 0$

So, (x-3) is not a factor

Now again hit and trail method

taking value of x as -1

$\tt p(-1)=(-1)^4+(-1)^3-7(-1)^2-(-1)+6 \\\\ \tt =1-1-7+1+6 \\\\ \tt =0$

So, x= -1 is a zero

So, (x+1) is a factor

Now again applying hit and trial method

Taking x = -3

$\tt p(-3)=(-3)^4+(-3)^3-7(-3)^2-(-3)+6 \\\\ \tt =81-27-63+3+6 \\\\ \tt = 90-90 \\\\ \tt =0$

So, x=-3 is a zero

so, (x+3) is a factor

So, factors of x^4 + x^3 - 7x^2 -x + 6 are (x+1),(x-1),(x-2) and (x+3)