Factorise? a) x^4+2x^3+3x^2+2x+1

2 Answers
Oscar L.
Jan 4, 2018

#x^4+2x^3+3x^2+2x+1=(x^2+x+1)^2#.

Explanation:

When you have a "symmetric polynomial", one where the coefficients are the same when you read them forwards and backwards, like #x^4+2x^3+3x^2+2x+1#, then it will have symmetric factors too.

Let us assume that the given fourth degree polynomial has a pair of symmetric, quadratic factors. The natural choice, since the first and last coefficients are both #1#, is:

#x^4+2x^3+3x^2+2x+1=(x^2+ax+1)(x^2+bx+1)#

Multiply the factors on the right:

#x^4+2x^3+3x^2+2x+1=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1#

So

#a+b=2#, thus #b=2-a#

#ab=a(2-1)=1#, thus #a^2-2a+1=0#

The only root for #a# is #a=1#, then also #b=1# and we end up with:

#x^4+2x^3+3x^2+2x+1=(x^2+x+1)^2#.

Ratnaker Mehta
Jan 4, 2018

# (x^2+x+1)^2#.

Explanation:

If we complete the square #x^4+2x^3#, we will get, #x^2# as the

the third term.

#:. x^4+2x^3+color(red)(3x^2)+2x+1#,

#=(x^4+2x^3+color(red)(x^2))+ul(color(red)(2x^2)+2x)+1#,

#={(x^2)^2+2(x^2)(x)+(x)^2}+2(x^2+x)+1#,

#=(x^2+x)^2+2(x^2+x)(1)+1^2#,

#=y^2+2y+1," where, "y=x^2+x#,

#=(y+1)^2#.

Sub.ing #y=x^2+x,# we find,

#"The Expression="(x^2+x+1)^2#.

Enjoy Maths., and Spread the Joy!

Related questions
Impact of this question
16986 views around the world
You can reuse this answer
Creative Commons License