Factorise:
`2x ^(2) + x - 45`

Answer

To factorise the expression \(2x^2 + x - 45\), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic expression is \(2x^2 + x - 45\). Here, the coefficients are:
- Coefficient of \(x^2\) (a) = 2
- Coefficient of \(x\) (b) = 1
- Constant term (c) = -45

### Step 2: Multiply the coefficient of \(x^2\) by the constant term
Next, we multiply the coefficient of \(x^2\) (which is 2) by the constant term (which is -45):
\[ 2 \times -45 = -90 \]

### Step 3: Find two numbers that multiply to -90 and add to the coefficient of \(x\)
We need to find two numbers that multiply to -90 and add to the coefficient of \(x\) (which is 1). The two numbers that satisfy this condition are 10 and -9:
\[ 10 \times -9 = -90 \quad \text{and} \quad 10 + (-9) = 1 \]

### Step 4: Rewrite the middle term using the two numbers
Now, we can rewrite the expression \(2x^2 + x - 45\) by splitting the middle term (x) using the two numbers we found:
\[ 2x^2 + 10x - 9x - 45 \]

### Step 5: Group the terms
Next, we group the terms in pairs:
\[ (2x^2 + 10x) + (-9x - 45) \]

### Step 6: Factor out the common factors from each group
Now, we factor out the common factors from each group:
\[ 2x(x + 5) - 9(x + 5) \]

### Step 7: Factor out the common binomial factor
Notice that \((x + 5)\) is a common factor:
\[ (2x - 9)(x + 5) \]

### Final Result
Thus, the factorised form of \(2x^2 + x - 45\) is:
\[ (2x - 9)(x + 5) \]

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