Question

X^2(x^2+3x+2)plzz help me

Answer 1

The zeros for this function are -2, -1 and a double root of 0.

You can find this by first factoring the polynomial on the inside of the parenthesis. Polynomials like this can be factored by looking for two numbers that multiply to the constant (2) and add up to the second coefficient (3). The numbers 2 and 1 satisfy both of those needs and thus can be used as the numbers in a factoring.

x^2(x^2 + 3x + 2)

x^2(x + 2)(x + 1)

Now to find the zeros, we set each part equal to 0. You may want to split the x^2 into two separate x's for this purpose.

(x)(x)(x + 2)(x + 1)

x = 0

x = 0

x + 2 = 0

x = -2

x + 1 = 0

x = -1

Answer 2

Distribute the x^2 into all the terms of the parenthesis:

x^2 * x^2 = x^4 (When you have exponents in a equation that looks like this you always add the exponents together)

x^2 * 3x = 3x^3 (3x still has an exponent of 1 so you add that plus 2)

x^2 * 2 = 2x^2 (there is no exponent in the term of 2 because there is no x so you just add the x^2 onto the 2)

So you're simplified expression is x^4 + 3x^3 + 2x^2

Hope this helped! Mark Brainliest please! :)))