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.
