Factorise the algebraic expressions by finding the common factor. \( \begin{array}{ll}\text { a) } y^{2}+8 y & \text { b) } a b c-a c \\ \text { c) } 3 a^{2}-12 a & \text { d) } x y+y-y z\end{array} \) e) \( -4 b-20 \) g) \( (x-1)-y(x-1) \) Factorise the expressions consisting of the difference of two square Test the answers by multiplying. \( \begin{array}{ll}\text { a) } x^{2}-y^{2} & \text { h) } k(m-n)-15 a^{2} b+a b \\ \text { c) } m^{2}-4 \cdot n^{2} & \text { b) } 9 x^{2}-y^{2} \\ \text { e) } x^{4}-y^{6} & \left.\text { d) } 16 a^{2}-25 \cdot\right)^{2} \\ \text { g) } 100-m^{2} n^{2} & \text { f) }(x-y)^{2}-\{a\end{array} \)
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Bonus Knowledge
To factor the given algebraic expressions, let’s dive into some common factors! For \(y^{2}+8y\), the common factor is \(y\), which gives us \(y(y + 8)\). In the case of \(abc - ac\), the common factor \(ac\) allows us to factor it as \(ac(b - 1)\). Likewise, \(3a^{2}-12a\) can be factored to \(3a(a - 4)\). Not to forget \(xy + y - yz\); factoring out \(y\) yields \(y(x + 1 - z)\). Lastly, for \(-4b - 20\), we see that \(-4\) is a common factor, resulting in \(-4(b + 5)\). Next, let's tackle the difference of squares! The expression \(x^{2}-y^{2}\) can be factored into \((x+y)(x-y)\). For \(9x^{2} - y^{2}\), we utilize the same concept to get \((3x + y)(3x - y)\). Don’t forget \(m^{2}-4n^{2}\)—this can also be factored using the difference of squares as \((m - 2n)(m + 2n)\). The expression \(16a^{2} - 25\) simplifies to \((4a - 5)(4a + 5)\). Lastly, \(100 - m^{2}n^{2}\) can be written as \((10 - mn)(10 + mn)\). Don't forget to check your answers by multiplying the factors back together—it's like a magic trick where the outcome reveals the original expression!
