Fuctorise \( x-6 x^{\frac{1}{2}} y+8 y^{2} \)
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To factor the expression \( x - 6 x^{\frac{1}{2}} y + 8 y^{2} \), we can first treat \( x^{\frac{1}{2}} \) as a new variable. Let's set \( z = x^{\frac{1}{2}} \), then \( z^2 = x \). Substituting that in gives: \( z^2 - 6zy + 8y^2 \). Now, we need to factor this quadratic expression. We look for two numbers that multiply to \( 8 \) (the constant term) and add to \( -6 \) (the coefficient of \( zy \)). Those numbers are \( -4 \) and \( -2 \). Thus, we can factor the expression to: \( (z - 4y)(z - 2y) \). Substituting back \( z = x^{\frac{1}{2}} \): \( (x^{\frac{1}{2}} - 4y)(x^{\frac{1}{2}} - 2y) \). So the factorization of the original expression \( x - 6 x^{\frac{1}{2}} y + 8 y^{2} \) is \( (x^{\frac{1}{2}} - 4y)(x^{\frac{1}{2}} - 2y) \).