$$ \begin{array}{ll} ext { 2.1.1 } & (2 x-3)(2 x+3) \ ext { 2.1.2 } & \left(x^{2}+y ight)\left(4 x+3 x y+x^{-1} y^{2} ight) \ ext { 2.1.3 } & \frac{x+1}{2}-\frac{2-3 x}{3}+x \ & ext { Factorise each of the following expressions: } \ ext { 2.2.1 } & 20 x^{2}-45 y^{2} \ ext { 2.2.2 } & 2 x^{2}-5 x+3 \ ext { 2.2.3 } & 5 x^{2}-15 x+12 y-4 x y \ & ext { Solve for } x: \ ext { 2.3.1 } & 27^{x+1}=9^{x-2} \ ext { 2.3.2 } & \frac{2}{x}=\frac{-4}{x+5} \ & \end{array} $$

Question

$$ \begin{array}{ll}	ext { 2.1.1 } & (2 x-3)(2 x+3) \ 	ext { 2.1.2 } & \left(x^{2}+yight)\left(4 x+3 x y+x^{-1} y^{2}ight) \ 	ext { 2.1.3 } & \frac{x+1}{2}-\frac{2-3 x}{3}+x \ & 	ext { Factorise each of the following expressions: } \ 	ext { 2.2.1 } & 20 x^{2}-45 y^{2} \ 	ext { 2.2.2 } & 2 x^{2}-5 x+3 \ 	ext { 2.2.3 } & 5 x^{2}-15 x+12 y-4 x y \ & 	ext { Solve for } x: \ 	ext { 2.3.1 } & 27^{x+1}=9^{x-2} \ 	ext { 2.3.2 } & \frac{2}{x}=\frac{-4}{x+5} \ & \end{array} $$

UpStudy AI · Accepted Answer

### 2.1.1  
We simplify  
$$
(2x-3)(2x+3)
$$  
by recognizing it as a product of conjugates. Using the formula  
$$
(a-b)(a+b)=a^2-b^2,
$$  
with $$a=2x$$ and $$b=3$$, we obtain  
$$
(2x-3)(2x+3)=(2x)^2-3^2=4x^2-9.
$$

---

### 2.1.2  
We expand  
$$
\left(x^2+y\right)\left(4x+3xy+x^{-1}y^2\right).
$$  
Multiply each term in the first bracket by each term in the second bracket:

- $$x^2\cdot 4x=4x^3$$,
- $$x^2\cdot 3xy=3x^3y$$,
- $$x^2\cdot x^{-1}y^2=x^{2-1}y^2=x y^2$$,
- $$y\cdot 4x=4xy$$,
- $$y\cdot 3xy=3xy^2$$,
- $$y\cdot x^{-1}y^2=x^{-1}y^3$$.

Collecting these terms gives
$$
4x^3+3x^3y+xy^2+4xy+3xy^2+x^{-1}y^3.
$$
Notice that $$xy^2$$ and $$3xy^2$$ can be combined:
$$
xy^2+3xy^2=4xy^2.
$$
Thus, the expanded expression is
$$
4x^3+3x^3y+4xy+4xy^2+x^{-1}y^3.
$$

---

### 2.1.3  
We simplify  
$$
\frac{x+1}{2}-\frac{2-3x}{3}+x.
$$  
First, find a common denominator. The common denominator for 2, 3, and (implicitly) 1 is 6. Rewrite each term:

- $$\frac{x+1}{2}=\frac{3(x+1)}{6}=\frac{3x+3}{6}$$,
- $$\frac{2-3x}{3}=\frac{2(2-3x)}{6}=\frac{4-6x}{6}$$ (note the minus sign in front gives $$-\frac{4-6x}{6}=\frac{-4+6x}{6}$$),
- $$x=\frac{6x}{6}$$.

Now, combine the fractions:
$$
\frac{3x+3+(-4+6x)+6x}{6}=\frac{3x+3-4+6x+6x}{6}=\frac{15x-1}{6}.
$$

---

### 2.2.1  
We factorise  
$$
20x^2-45y^2.
$$  
First, factor out the common factor 5:
$$
20x^2-45y^2=5(4x^2-9y^2).
$$
Recognize $$4x^2-9y^2$$ as a difference of squares:
$$
4x^2-9y^2=(2x)^2-(3y)^2=(2x-3y)(2x+3y).
$$
Thus, the fully factorised form is:
$$
5(2x-3y)(2x+3y).
$$

---

### 2.2.2  
We factorise  
$$
2x^2-5x+3.
$$  
We look for two numbers which multiply to $$2\times3=6$$ and add to $$-5$$. The numbers $$-2$$ and $$-3$$ work since $$-2\times -3=6$$ and $$-2+(-3)=-5$$.

Rewrite the middle term:
$$
2x^2-2x-3x+3.
$$
Group the terms:
$$
(2x^2-2x)- (3x-3)=2x(x-1)-3(x-1).
$$
Factor out the common factor $$(x-1)$$:
$$
(2x-3)(x-1).
$$

---

### 2.2.3  
We factorise  
$$
5x^2-15x+12y-4xy.
$$  
Group the terms into two groups:
$$
(5x^2-15x)+(-4xy+12y).
$$
Factor out common factors from each group:
- From $$5x^2-15x$$, factor out $$5x$$: $$5x(x-3)$$.
- From $$-4xy+12y$$, factor out $$-4y$$: $$-4y(x-3)$$.

Now, factor out the common factor $$(x-3)$$:
$$
(x-3)(5x-4y).
$$

---

### 2.3.1  
We solve for $$x$$ in the exponential equation:
$$
27^{x+1}=9^{x-2}.
$$
Express 27 and 9 in terms of 3:
$$
27=3^3 \quad \text{and} \quad 9=3^2.
$$
Thus, rewrite the equation:
$$
(3^3)^{x+1}=(3^2)^{x-2}.
$$
Simplify by multiplying exponents:
$$
3^{3(x+1)}=3^{2(x-2)}.
$$
Since the bases are the same, set the exponents equal:
$$
3(x+1)=2(x-2).
$$
Solve for $$x$$:
$$
3x+3=2x-4 \quad \Rightarrow \quad 3x-2x=-4-3 \quad \Rightarrow \quad x=-7.
$$

---

### 2.3.2  
We solve for $$x$$ in the equation:
$$
\frac{2}{x}=\frac{-4}{x+5}.
$$
Cross-multiply:
$$
2(x+5)=-4x.
$$
Expand the left-hand side:
$$
2x+10=-4x.
$$
Add $$4x$$ to both sides:
$$
6x+10=0.
$$
Subtract 10 from both sides:
$$
6x=-10.
$$
Divide both sides by 6:
$$
x=-\frac{10}{6}=-\frac{5}{3}.
$$
**2.1.1** $$ (2x-3)(2x+3) = 4x^2 - 9 $$ **2.1.2** $$ (x^2 + y)(4x + 3xy + x^{-1}y^2) = 4x^3 + 3x^3y + xy^2 + 4xy + 3xy^2 + x^{-1}y^3 $$ **2.1.3** $$ \frac{x+1}{2} - \frac{2-3x}{3} + x = \frac{15x - 1}{6} $$ **2.2.1** $$ 20x^2 - 45y^2 = 5(2x - 3y)(2x + 3y) $$ **2.2.2** $$ 2x^2 - 5x + 3 = (2x - 3)(x - 1) $$ **2.2.3** $$ 5x^2 - 15x + 12y - 4xy = (x - 3)(5x - 4y) $$ **2.3.1** $$ x = -7 $$ **2.3.2** $$ x = -\frac{5}{3} $$

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Algebra Questions