What is the standard form of y= x^2(x-9) (6-x)y=x2(x−9)(6−x)?

1 Answer
Shwetank Mauria
Jun 30, 2016

y=-x^4+15x^3-54x^2y=−x4+15x3−54x2

Explanation:

In y=x^2(x-9)(6-x)y=x2(x−9)(6−x), the RHS is a polynomial of degree 44 in xx, as xx gets multiplied four times.

The standard form of a polynomial in degree 44 is ax^4+bx^3+cx^2+dx+fax4+bx3+cx2+dx+f, for which we should expand x^2(x-9)(6-x)x2(x−9)(6−x) by multiplying.

x^2(x-9)(6-x)x2(x−9)(6−x)

= x^2(x(6-x)-9(6-x))x2(x(6−x)−9(6−x))

= x^2(6x-x^2-54+9x)x2(6x−x2−54+9x)

= x^2(-x^2+15x-54)x2(−x2+15x−54)

= -x^4+15x^3-54x^2−x4+15x3−54x2

Note that here coefficient of xx and constant terms are both zero in this case.

Related questions
See all questions in Polynomials in Standard Form
Impact of this question
1432 views around the world
You can reuse this answer
Creative Commons License