Factorise 1.) \( 4 x-x^{3} \) 2.) \( x^{2}+5 x+4 \) 3.) \( x-x y+x-1 \)
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To factorize \( 4x - x^{3} \), you can start by taking out the common factor. Notice that both terms have \( x \) as a common factor. Thus, you can write it as \( x(4 - x^{2}) \). The expression \( 4 - x^{2} \) can be further factored using the difference of squares: \( 4 - x^{2} = (2 - x)(2 + x) \). So the complete factorization is \( x(2 - x)(2 + x) \). For \( x^{2} + 5x + 4 \), you're looking for two numbers that multiply to 4 (the constant term) and add up to 5 (the coefficient of \( x \)). The numbers 1 and 4 fit the bill! Therefore, you can factor it as \( (x + 1)(x + 4) \). Next, for \( x - xy + x - 1 \), you can first rearrange and combine like terms: \( 2x - xy - 1 \). Now, observe that \( 2x - xy \) can be factored as \( x(2 - y) \). Thus, you can rewrite the whole expression as \( x(2 - y) - 1 \). This expression does not factor neatly into simpler binomials; however, it can be expressed as a combination of those terms. So, the final factorizations are: 1.) \( x(2 - x)(2 + x) \) 2.) \( (x + 1)(x + 4) \) 3.) \( x(2 - y) - 1 \) (though it may not factor further, this is a simplified form).
