Jom Cuba 2.1 (Soalan 1 – 3) – Buku Teks Matematik Tingkatan 2 Bab 2 (Pemfaktoran dan Pecahan Algebra)


Soalan 1:
Berdasarkan jubin algebra berikut, tulis luas kawasan berlorek dalam bentuk pendaraban dua ungkapan algebra.



Penyelesaian:
(a)
$$ \begin{aligned} \text { Luas kawasan berlorek } & =\text { panjang } \times \text { lebar } \\ & =(a+1+1)(a+1) \\ & =(a+2)(a+1) \end{aligned} $$
(b)
$$ \begin{aligned} \text { Luas kawasan berlorek } & =\text { panjang } \times \text { lebar } \\ & =(4 x-3)(4 x-3) \end{aligned} $$


Soalan 2:
Kembangkan ungkapan algebra berikut.
(a) 3(x + 2)
(b) 4(8x − 3)
(c) 2(a + 5)
(d) p(6p − 8)
(e) − r/8 (2s − 8)
(f) −2(pr − 2pq)
(g) 3(5bc − 6)
(h) 7(2ef + 3e)
(i) 8g(2 + gh)

Penyelesaian:
(a)
$$ \begin{aligned} & 3(x+2) \\ & =(3 \times x)+(3 \times 2) \\ & =3 x+6 \end{aligned} $$
(b)
$$ \begin{aligned} & 4(8 x-3) \\ & =(4 \times 8 x)+(4 \times-3) \\ & =32 x-12 \end{aligned} $$


(c)
$$ \begin{aligned} & 2(a+5) \\ & =(2 \times a)+(2 \times 5) \\ & =2 a+10 \end{aligned} $$
(d)
$$ \begin{aligned} & p(6 p-8) \\ & =(p \times 6 p)+(p \times-8) \\ & =6 p^2-8 p \end{aligned} $$


(e)
$$ \begin{aligned} & -\frac{r}{8}(2 s-8) \\ = & \left(-\frac{r}{8} \times 2 s\right)+\left(-\frac{r}{8} \times-8\right) \\ = & -\frac{r s}{4}+r \end{aligned} $$
(f)
$$ \begin{aligned} & -2(p r-2 p q) \\ & =(-2 \times p r)+(-2 \times-2 p q) \\ & =-2 p r+4 p q \end{aligned} $$


(g)
$$ \begin{aligned} & 3(5 b c-6) \\ & =(3 \times 5 b c)+(3 \times-6) \\ & =15 b c-18 \end{aligned} $$
(h)
$$ \begin{aligned} & 7(2 e f+3 e) \\ & =(7 \times 2 e f)+(7 \times 3 e) \\ & =14 e f+21 e \end{aligned} $$


(i)
$$ \begin{aligned} & 8 g(2+g h) \\ & =(8 g \times 2)+(8 g \times g h) \\ & =16 g+8 g^2 h \end{aligned} $$


Soalan 3:
Kembangkan ungkapan algebra berikut.
(a) (a + 1)(a + 2)
(b) (x − 5)(x + 4)
(c) (2 + m)(5 − m)
(d) (3p − 2)(4p − 1)
(e) (3r − 2)(4r − 1)
(f) (2r + s)(4r − 3s)
(g) (2d − ½b)(3d − ½b)
(h) (r − 3s)2
(i) (4e − 3)2

Penyelesaian:
(a)
$$ \begin{aligned} &(a+1)(a+2)\\ &\begin{aligned} & =a(a+2)+1(a+2) \\ & =a^2+2 a+a+2 \\ & =a^2+3 a+2 \end{aligned} \end{aligned} $$
(b)
$$ \begin{aligned} &(x-5)(x+4)\\ &\begin{aligned} & =x(x+4)-5(x+4) \\ & =x^2+4 x-5 x-20 \\ & =x^2-x-20 \end{aligned} \end{aligned} $$


(c)
$$ \begin{aligned} &(2+m)(5-m)\\ &\begin{aligned} & =2(5-m)+m(5-m) \\ & =10-2 m+5 m-m^2 \\ & =10+3 m-m^2 \end{aligned} \end{aligned} $$
(d)
$$ \begin{aligned} & (3 p-2)(4 p-1) \\ & =3 p(4 p-1)-2(4 p-1) \\ & =12 p^2-3 p-8 p+2 \\ & =12 p^2-11 p+2 \end{aligned} $$


(e)
$$ \begin{aligned} & (3 r-2)(4 r-1) \\ & =3 r(4 r-1)-2(4 r-1) \\ & =12 r^2-3 r-8 r+2 \\ & =12 r^2-11 r+2 \end{aligned} $$
(f)
$$ \begin{aligned} &(2 r+s)(4 r-3 s)\\ &\begin{aligned} & =2 r(4 r-3 s)+s(4 r-3 s) \\ & =8 r^2-6 r s+4 r s-3 s^2 \\ & =8 r^2-3 s^2-2 r s \end{aligned} \end{aligned} $$


(g)
$$ \begin{aligned} & \left(2 d-\frac{1}{2} b\right)\left(3 d-\frac{1}{2} b\right) \\ & =2 d\left(3 d-\frac{1}{2} b\right)-\frac{1}{2} b\left(3 d-\frac{1}{2} b\right) \\ & =6 d^2-b d-\frac{3}{2} b d+\frac{1}{4} b^2 \\ & =6 d^2-\frac{5}{2} b d+\frac{1}{4} b^2 \end{aligned} $$


(h)
$$ \begin{aligned} & (r-3 s)^2 \\ & =(r-3 s)(r-3 s) \\ & =r^2-3 r s-3 r s+9 s^2 \\ & =r^2-6 r s+9 s^2 \end{aligned} $$


(i)
$$ \begin{aligned} & (4 e-3)^2 \\ & =(4 e-3)(4 e-3) \\ & =16 e^2-12 e-12 e+9 \\ & =16 e^2-24 e+9 \end{aligned} $$

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