Question

1. $x^2 - 16xy + 28y^2$

Explanation:

Step1: Identify the form

The quadratic expression is \( x^2 - 16xy + 28y^2 \), which is of the form \( ax^2+bx y + cy^2 \) with \( a = 1 \), \( b=- 16y \), \( c = 28y^2 \) (treating \( y \) as a constant for factoring in terms of \( x \)). We need two numbers that multiply to \( a\times c=1\times28y^2 = 28y^2 \) and add up to \( b=-16y \).

Step2: Find the factors

We need two numbers \( m \) and \( n \) such that \( m\times n=28y^2 \) and \( m + n=-16y \). The factors of \( 28y^2 \) that add up to \( - 16y \) are \( -2y \) and \( -14y \) since \( (-2y)\times(-14y)=28y^2 \) and \( (-2y)+(-14y)=-16y \).

Step3: Split the middle term

Rewrite the middle term using these factors: \( x^2-2xy - 14xy + 28y^2 \)

Step4: Factor by grouping

Group the terms: \( (x^2 - 2xy)+(-14xy + 28y^2) \)

Factor out the greatest common factor from each group: \( x(x - 2y)-14y(x - 2y) \)

Step5: Factor out the common binomial factor

Now, factor out \( (x - 2y) \) from both terms: \( (x - 2y)(x - 14y) \)

Answer:

\( (x - 2y)(x - 14y) \)