Question

(08.01, 08.02, 08.03, 08.05, 08.06 MC)

Part A: Factor x2y2 + 6xy2 + 8y2. Show your work. (4 points)

Part B: Factor x2 + 8x + 16. Show your work. (3 points)

Part C: Factor x2 − 16. Show your work. (3 points)

Answer 1

is it x times 2 times y times 2? is that what it is asking or is it meaning squared

Answer 2

Answer: The factorization of all the parts are :

Part A : $x^2y^2+6xy^2+8y^2=y^2(x+2)(x+4).$

Part B :  $x^2+8x+16=(x+4)(x+4).$

Part C :  $x^2-16=(x+4)(x-4).$

Step-by-step explanation:  We are given to factorize the following quadratic polynomials :

Part A : Factor x²y² + 6xy² + 8y².  

Part B : Factor x² + 8x + 16.

Part C: Factor x² − 16.

We will be using the following factorization formulas :

$(i)~x^2+ax+bx+ab=(x+a)(x+b),\\\\(ii)~x^2+2xa+a^2=(x+a)^2=(x+a)(x+a),\\\\(iii)~x^2-a^2=(x+a)(x-a).$

The factorization of all the parts area as follows :

Part A :

We have

$x^2y^2+6xy^2+8y^2\\\\=y^2(x^2+6x+8)\\\\=y^2(x^2+4x+2x+8)\\\\=y^2(x(x+4)+2(x+4))\\\\=y^2(x+2)(x+4).$

So, $x^2y^2+6xy^2+8y^2=y^2(x+2)(x+4).$

Part B :

We have

$x^2+8x+16\\\\=x^2+2\times x\times4+4^2\\\\=(x+4)^2\\\\=(x+4)(x+4).$

So, $x^2+8x+16=(x+4)(x+4).$

Part C :

We have

$x^2-16\\\\=x^2-4^2\\\\=(x+4)(x-4).$

So, $x^2-16=(x+4)(x-4).$

Thus, all the parts are factorized.