The factorisation of x 2+ 11xy + 24y 2is:
(x + 8y) (x + 3y)
Factorising Quadratic Trinomials like \(x^2 + 11xy + 24y^2\)
We are asked to factor the expression \(x^2 + 11xy + 24y^2\). This is a quadratic trinomial with two variables, x and y. It is in the standard form \(ax^2 + bxy + cy^2\), where \(a=1\), \(b=11\), and \(c=24\).
To factor this type of trinomial when \(a=1\), we look for two numbers that multiply to \(c\) (the coefficient of \(y^2\), which is 24) and add up to \(b\) (the coefficient of \(xy\), which is 11).
Let the two numbers be \(p\) and \(q\). We need:
- \(p \times q = 24\)
- \(p + q = 11\)
Let's list the pairs of factors of 24 and check their sums:
| Pair of Factors of 24 | Sum of Factors |
|---|---|
| 1, 24 | 1 + 24 = 25 |
| 2, 12 | 2 + 12 = 14 |
| 3, 8 | 3 + 8 = 11 |
| 4, 6 | 4 + 6 = 10 |
The pair of factors that adds up to 11 is 3 and 8.
Now we can rewrite the middle term, \(11xy\), using these two numbers: \(11xy = 3xy + 8xy\).
Substitute this back into the original expression:
\(x^2 + 11xy + 24y^2 = x^2 + 3xy + 8xy + 24y^2\)
Next, we factor by grouping the terms:
Group the first two terms and the last two terms:
\((x^2 + 3xy) + (8xy + 24y^2)\)
Factor out the common factor from each group:
- From \((x^2 + 3xy)\), the common factor is \(x\). Factoring it out gives \(x(x + 3y)\).
- From \((8xy + 24y^2)\), the common factor is \(8y\). Factoring it out gives \(8y(x + 3y)\).
So the expression becomes:
\(x(x + 3y) + 8y(x + 3y)\)
Now, we can see that \((x + 3y)\) is a common binomial factor in both terms. Factor out \((x + 3y)\):
\((x + 3y)(x + 8y)\)
This is the factorised form of the expression \(x^2 + 11xy + 24y^2\).
Let's compare this with the given options:
- Option 1: \((x - 8y) (x + 3y)\)
- Option 2: \((x + 8y) (x - 3y)\)
- Option 3: \((x + 8y) (x + 3y)\)
- Option 4: \((x - 8y) (x - 3y)\)
Our result, \((x + 3y)(x + 8y)\), is the same as \((x + 8y)(x + 3y)\) because multiplication is commutative. This matches Option 3.
Revision Table: Key Concepts for Factorisation
| Concept | Description | Example |
|---|---|---|
| Factorisation | Breaking down an expression into a product of simpler expressions (factors). | \(x^2 - 4 = (x-2)(x+2)\) |
| Trinomial | A polynomial with three terms. | \(ax^2 + bx + c\) or \(x^2 + 11xy + 24y^2\) |
| Quadratic Trinomial | A trinomial where the highest power of the variable(s) is 2. | \(x^2 + 5x + 6\) |
| Factoring \(ax^2 + bxy + cy^2\) when \(a=1\) | Find two numbers \(p, q\) such that \(p+q=b\) and \(pq=c\), then write as \((x+py)(x+qy)\). | For \(x^2 + 5xy + 6y^2\), find \(p,q\) such that \(p+q=5, pq=6\). Numbers are 2, 3. Factors are \((x+2y)(x+3y)\). |
Additional Information: Factoring Quadratic Expressions
Factoring quadratic expressions is a fundamental skill in algebra. The method used here, often called splitting the middle term, is applicable to trinomials of the form \(ax^2 + bx + c\) or \(ax^2 + bxy + cy^2\).
Steps for Factoring \(ax^2 + bxy + cy^2\) by Splitting the Middle Term:
- Identify \(a\), \(b\), and \(c\).
- Calculate the product \(ac\).
- Find two numbers \(p\) and \(q\) such that \(p+q = b\) and \(pq = ac\).
- Rewrite the middle term \(bxy\) as \(pxy + qxy\).
- Group the terms: \((ax^2 + pxy) + (qxy + cy^2)\).
- Factor out the greatest common divisor (GCD) from each group. You should end up with a common binomial factor.
- Factor out the common binomial factor.
In our problem, \(a=1\), \(b=11\), \(c=24\). We needed \(p+q=11\) and \(pq=1 \times 24 = 24\). We found \(p=3\) and \(q=8\). The expression became \(x^2 + 3xy + 8xy + 24y^2\), which factored to \((x+3y)(x+8y)\).
Understanding how to find the correct pair of numbers is crucial. Always consider both positive and negative factors depending on the signs of \(b\) and \(c\).
Important Questions from Polynomials
In the expansion of (x + 3) 3, the coefficient of x is:
If the degree of polynomial 9 x 5y 2z ris 15, then r = ?
The value of 16x 4+ 25y 2– 40x 2y at x = 5 and y = 2 is:
If the sum of the squares of the zeros of quadratic polynomial f(x) = x 2– 8x + k is 40, then find the value of k.
Expand : (s + 2) 3
A. s 3+ 2s 2+ 12s + 8
B. s 3+ 3s 2+ 6s + 8
C. s 3+ 6s 2+ 12s + 8
D. s 3+ 6s 2+ 6s + 8
