1 \).
- The coefficient of \( x \) (middle term) is \( 6 \).
- The constant term (last term) is \( 8 \).
### Step 2: Calculate the product of the first and last term
We need to find the product of the coefficient of \( x^2 \) and the constant term:
\[
1 \times 8 = 8
\]
### Step 3: Find two numbers that add up to the middle term and multiply to the product
We need to find two numbers that add up to \( 6 \) (the coefficient of \( x \)) and multiply to \( 8 \) (the product we calculated). The numbers \( 4 \) and \( 2 \) satisfy this:
\[
4 + 2 = 6 \quad \text{and} \quad 4 \times 2 = 8
\]
### Step 4: Rewrite the middle term
Now, we can rewrite the expression by splitting the middle term \( 6x \) into \( 4x + 2x \):
\[
x^2 + 4x + 2x + 8
\]
### Step 5: Group the terms
Next, we group the terms:
\[
(x^2 + 4x) + (2x + 8)
\]
### Step 6: Factor out the common factors from each group
From the first group \( (x^2 + 4x) \), we can factor out \( x \):
\[
x(x + 4)
\]
From the second group \( (2x + 8) \), we can factor out \( 2 \):
\[
2(x + 4)
\]
### Step 7: Combine the factored terms
Now we have:
\[
x(x + 4) + 2(x + 4)
\]
We can see that \( (x + 4) \) is common in both terms, so we factor it out:
\[
(x + 4)(x + 2)
\]
### Final Result
Thus, the factorised form of the expression \( x^2 + 6x + 8 \) is:
\[
(x + 4)(x + 2)
\]
---

Factorise : `x^2 + 6x + 8`

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ICSE-FACTORISATION-EXERCISE 13(D)