Question

using factor theorem,factorise the polynomial u4+u3-7u2-u+6​

Answer 1

Answer:

(U+1)(U-1)(U-2)(U+3)

Step-by-step explanation:

let the value of u be 1,

p(u)=u4+u3-7u2-u+6

so,p(1)=(1)⁴+(1)³-7(1)²-(1)+6

= 1+1-7-1+6

=0

as p(1)=0, 1 is the root of p(u)

(u-1)is a factor of p(u)

now,let value of u be -1,

p(-1)=(-1)⁴+(-1)³-7(-1)²-(-1)+6

= 1-1-7+1+6

=0

as p(-1)=0,-1 is the root of p(u)

(u+1) is also one of the factor....

similarly by assuming the value of u as 2,we get remainder 0,

so (u-2) is again a factor...

even value of u= -3 gives remainder 0,

so (u+3)is also a factor.

hence,

u⁴+u³-7u²-u+6= (u+1)(u-1)(u-2)(u+3)