Question

Which polynomial is factored completely?

Answer 1

The completely factored polynomials are $\boxed{2x\left( {{x^2} - 4} \right)}$ and $\boxed{\left( {4x + 4} \right)\left( {x + 1} \right)}$. Option (b) and option (c) are correct.

Further Explanation:

Given:

The options are as follows,

(a). $121{x^2} + 36{y^2}$

(b). $\left( {4x + 4} \right)\left( {x + 1} \right)$

(c). $3{x^4} - 15{n^3} + 12{n^2}$

(d). $3{x^4} - 15{n^3} + 12{n^2}$

Calculation:

In option (a)

The expression $121{x^2} + 36{y^2}$ is not in the factored form. option (a) is not correct.

In option (b)

The expression $\left( {4x + 4} \right)\left( {x + 1} \right)$ is in the factored form.

${\text{Factored}} = 4{\left( {x + 1} \right)^2}$

Option (b) is correct.

In option (c)

The expression $2x\left( {{x^2} - 4} \right)$ is in the factored form.

$\begin{aligned}{\text{Factored}} &= 2x\left( {{x^2} - 4} \right)\\&= 2x\left( {x + 2} \right)\left( {x - 2} \right)\\\end{aligned}$

Option (c) is correct.

In option (d)

The expression $3{x^4} - 15{n^3} + 12{n^2}$ is not in the factored form.

Option (d) is not correct.

The completely factored polynomials are $\boxed{2x\left( {{x^2} - 4} \right)}$ and $\boxed{\left( {4x + 4} \right)\left( {x + 1} \right)}.$ Option (b) and option (c) are correct.

Learn more:

1. Learn more about unit conversion brainly.com/question/4837736
2. Learn more about non-collinear brainly.com/question/4165000
3. Learn more aboutbinomial and trinomial brainly.com/question/1394854  

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Exponents and Powers

Keywords: solution, factored, completely, polynomial, factorized form, expression, difference of cubes, exponents, power, equation, power rule, exponent rule.

Answer 2

Answer:

Complete factored: (4x+4)(x+1) and 2x(x²-4)

Step-by-step explanation:

We have to check each option one by one

Option 1: $121x^2+36y^2$

It is sum of two perfect square, As we know sum of two perfect square can't be factored. It is prime equation.

False

Option 2: $(4x+4)(x+1)$

Here we have two factor (4x+4) and (x+1). In both factor degree of x is 1. It is complete factor form.

Complete factor: $4(x+1)(x+1)$

True

Option 3: $2x(x^2-4)$

Here we have two factor 4x and x²-4. For second factor it is further factor. Because it is difference of square. Factor of x²-4 = (x+2)(x-2)

Complete Factor: 2x(x+2)(x-2)[/tex]

True

Option 4: $3x^4-15n^3+12n^2$

Here variable are not same. So, it can't be factor.

False