Question

X^3+3x^2+3x+1 how to factorise this with factor theorem

Answer 1

Answer:

(x+1)^3

Step-by-step explanation:

using the rational root theorem give

P=1. q=1

All value of +- p/q are - 1 & 1

Now, plug in each value to the polynomial by the remainder theram if the output is zero the input must be a root

(-1)^3+3(-1)^2+3(-1)+1

-1+3-3+1

their is a root x=-1

(1)^3+3(1)^2+3(1)+1

1+3+3+1=8

their is no root x=1

Factor of polynomial (x-root) =(x-(-1))

X+1

Now divided the polynomial by (x+1)

X^3+3x^2+3x+1/x+1

X^3+x^2+x^2+x+x^2+x+x+1/x+1

(x^3+x^2)+(x^2+x)+(x^2+x)+(x+1)/x+1

X^2+x+x+1

X^2+2x+1

You could do the same process with this new polynomials or you could recognize it as a perfect square

The result for this polynomial result in (x+1)^2

Answer 2

xx + 1 whole cube designs for p equal 1 q equal 1 + 3 by 2 or -1 and 1 - 1 whole cube + 3 into minus 1 square + 3 into minus 1 + 1 equal - 1 + 3 - 3 + 1 = 0 x = - 11 cube + 3 into 1 square + 3 into 1 + 1 equal 1 + 3 + 3 + 1 = 8 x = 1 x 1 x root equal x minus of minus 1 equal x + 1 divided by x + 1 x cube + 3 x square + 3 x + 1 divided by x + 1 x cube + x square + x square + x + x square + x + x + 1 / x + 1 x cube + x square + x square + x + x square + x + 6 + 1 / x + 1 x square + x + x + 1 equal x square + 2 x + 1 x + 10 square the 4 x + 1 into x + 1 whole square equal x + 1 whole cube this is the answer thank you