GSEB Class 8 Maths Solutions Chapter 14 Factorization Exercise 14.2

Question 1. Factorise the following expressions.
1. \( a^2 + 8a + 16 \)
2. \( p^2 - 10p + 25 \)
3. \( 25m^2 + 30m + 9 \)
4. \( 49y^2 + 84yz + 36z^2 \)
5. \( 4x^2 - 8x + 4 \)
6. \( 121b^2 - 88bc + 16c^2 \)
7. \( (l + m)^2 - 4lm \)
8. \( a^4 + 2a^2b^2 + b^4 \)
Answer:
1. We are given \( a^2 + 8a + 16 \). We can rewrite this as \( (a)^2 + 2(a)(4) + (4)^2 \). This expression matches the form of \( (x+y)^2 \), which is \( x^2 + 2xy + y^2 \). Thus, it simplifies to \( (a + 4)^2 \). We can also express this as the product of two identical factors: \( (a + 4)(a + 4) \).
In simple words: To factor this, we look for two numbers that multiply to 16 and add up to 8. Those numbers are 4 and 4, so it becomes \( (a+4)^2 \).
2. We are given \( p^2 - 10p + 25 \). This can be written as \( p^2 - 2(p)(5) + (5)^2 \). This matches the algebraic identity \( (x-y)^2 = x^2 - 2xy + y^2 \). So, it simplifies to \( (p - 5)^2 \). We can write this as two factors: \( (p - 5)(p - 5) \).
In simple words: For \( p^2 - 10p + 25 \), find two numbers that multiply to 25 and add up to -10. These numbers are -5 and -5. So, the expression factors into \( (p-5)^2 \).
3. We start with \( 25m^2 + 30m + 9 \). This can be expressed as \( (5m)^2 + 2(5m)(3) + (3)^2 \). This structure corresponds to the identity \( (x+y)^2 \). Therefore, it becomes \( (5m + 3)^2 \). This also means it is \( (5m + 3)(5m + 3) \).
In simple words: For \( 25m^2 + 30m + 9 \), look for a square root of \( 25m^2 \) and \( 9 \). These are \( 5m \) and \( 3 \). Check if twice their product is \( 30m \). Since \( 2 \times 5m \times 3 = 30m \), it is a perfect square trinomial \( (5m+3)^2 \).
4. We are given the expression \( 49y^2 + 84yz + 36z^2 \). This can be rewritten as \( (7y)^2 + 2(7y)(6z) + (6z)^2 \). This matches the perfect square formula \( (x+y)^2 \). Hence, it simplifies to \( (7y + 6z)^2 \). We can express this as the product \( (7y + 6z)(7y + 6z) \).
In simple words: For \( 49y^2 + 84yz + 36z^2 \), notice that \( 49y^2 \) is \( (7y)^2 \) and \( 36z^2 \) is \( (6z)^2 \). The middle term \( 84yz \) is \( 2 \times 7y \times 6z \). So, it's a perfect square and factors into \( (7y+6z)^2 \).
5. We begin with \( 4x^2 - 8x + 4 \). This expression is equivalent to \( (2x)^2 - 2(2x)(2) + (2)^2 \). This follows the form of \( (a-b)^2 \). Thus, it simplifies to \( (2x - 2)^2 \). We can also write this as \( (2x - 2)(2x - 2) \). By taking out a common factor of 2 from each bracket, we obtain \( 2(x-1) \times 2(x-1) \), which gives \( 4(x - 1)(x - 1) \).
In simple words: For \( 4x^2 - 8x + 4 \), we see that \( 4x^2 \) is \( (2x)^2 \) and \( 4 \) is \( (2)^2 \). The middle term \( -8x \) is \( -2 \times 2x \times 2 \). So, it is \( (2x-2)^2 \). We can factor out 2 from \( (2x-2) \) to get \( 2(x-1) \), so the final answer is \( 4(x-1)^2 \).
6. Given the expression \( 121b^2 - 88bc + 16c^2 \). This can be rewritten as \( (11b)^2 - 2(11b)(4c) + (4c)^2 \). This fits the pattern for \( (a-b)^2 \). Therefore, the expression simplifies to \( (11b - 4c)^2 \). This also means it is equal to \( (11b - 4c)(11b - 4c) \).
In simple words: For \( 121b^2 - 88bc + 16c^2 \), notice \( 121b^2 \) is \( (11b)^2 \) and \( 16c^2 \) is \( (4c)^2 \). The middle term \( -88bc \) is \( -2 \times 11b \times 4c \). So, it's a perfect square trinomial \( (11b-4c)^2 \).
7. We start with \( (l + m)^2 - 4lm \). First, we expand \( (l+m)^2 \) to get \( l^2 + 2lm + m^2 \). So the expression becomes \( l^2 + 2lm + m^2 - 4lm \). Now, we combine the like terms \( 2lm \) and \( -4lm \), which gives \( -2lm \). The expression simplifies to \( l^2 - 2lm + m^2 \). This is a perfect square trinomial, which can be written as \( (l - m)^2 \). This can also be stated as \( (l - m)(l - m) \).
In simple words: Expand \( (l+m)^2 \) to \( l^2+2lm+m^2 \). Then subtract \( 4lm \) from it. This leaves \( l^2-2lm+m^2 \), which is the formula for \( (l-m)^2 \).
8. We are given the expression \( a^4 + 2a^2b^2 + b^4 \). This can be rewritten by noticing that \( a^4 \) is \( (a^2)^2 \) and \( b^4 \) is \( (b^2)^2 \). So, the expression becomes \( (a^2)^2 + 2(a^2)(b^2) + (b^2)^2 \). This is a perfect square trinomial, matching the form \( (x+y)^2 \). Therefore, it simplifies to \( (a^2 + b^2)^2 \). This can also be written as the product of two factors: \( (a^2 + b^2)(a^2 + b^2) \).
In simple words: For \( a^4 + 2a^2b^2 + b^4 \), recognize that it's a perfect square. The square root of \( a^4 \) is \( a^2 \), and the square root of \( b^4 \) is \( b^2 \). The middle term is \( 2 \times a^2 \times b^2 \). So, it factors into \( (a^2+b^2)^2 \).

Exam Tip: Always double-check your factorization by expanding the result to ensure it matches the original expression.

Question 2. Factorise:
1. \( 4p^2 - 9q^2 \)
2. \( 63a^2 - 112b^2 \)
3. \( 49x^2 - 36 \)
4. \( 16x^5 - 144x^3 \)
5. \( (l + m)^2 - (l - m)^2 \)
6. \( 9x^2y^2 - 16 \)
7. \( (x^2 - 2xy + y^2) - z^2 \)
8. \( 25a^2 - 4b^2 + 28bc - 49c^2 \)
Answer:
1. We need to factorise \( 4p^2 - 9q^2 \). We can rewrite this as \( (2p)^2 - (3q)^2 \). This expression fits the difference of squares identity, \( a^2 - b^2 = (a - b)(a + b) \). Applying this rule, we get \( (2p - 3q)(2p + 3q) \).
In simple words: When you have two perfect squares subtracted, like \( A^2 - B^2 \), it factors into \( (A-B)(A+B) \). Here, \( A \) is \( 2p \) and \( B \) is \( 3q \).
2. We are given \( 63a^2 - 112b^2 \). First, we identify the greatest common factor (GCF). Both terms are divisible by 7. So we take out 7: \( 7(9a^2 - 16b^2) \). Inside the bracket, \( 9a^2 - 16b^2 \) is a difference of squares. We can write it as \( (3a)^2 - (4b)^2 \). Applying the formula \( a^2 - b^2 = (a + b)(a - b) \), this becomes \( (3a + 4b)(3a - 4b) \). So, the full factorization is \( 7(3a + 4b)(3a - 4b) \).
In simple words: First, take out the common number 7 from both parts. You'll be left with \( 9a^2 - 16b^2 \). This is a difference of two squares, which factors into \( (3a-4b)(3a+4b) \).
3. We need to factorise \( 49x^2 - 36 \). This expression can be rewritten as \( (7x)^2 - (6)^2 \). This is a difference of squares pattern. Using the identity \( a^2 - b^2 = (a - b)(a + b) \), we can factor it into \( (7x - 6)(7x + 6) \).
In simple words: For \( 49x^2 - 36 \), recognize that \( 49x^2 \) is \( (7x)^2 \) and \( 36 \) is \( (6)^2 \). It's a difference of squares, so the factors are \( (7x-6) \) and \( (7x+6) \).
4. Let's factorise \( 16x^5 - 144x^3 \). First, we find the greatest common factor (GCF). Both terms share \( 16x^3 \). So, we factor it out: \( 16x^3(x^2 - 9) \). Inside the bracket, \( x^2 - 9 \) is a difference of squares. We can write it as \( (x)^2 - (3)^2 \). Applying the identity \( a^2 - b^2 = (a + b)(a - b) \), this becomes \( (x + 3)(x - 3) \). Therefore, the fully factored expression is \( 16x^3(x + 3)(x - 3) \).
In simple words: Take out the common factor \( 16x^3 \). What's left inside is \( x^2 - 9 \). This is \( x^2 - 3^2 \), which factors into \( (x-3)(x+3) \).
5. We need to factorise \( (l + m)^2 - (l - m)^2 \). This expression is in the form of a difference of squares, \( A^2 - B^2 \), where \( A = (l+m) \) and \( B = (l-m) \). Using the identity \( A^2 - B^2 = (A + B)(A - B) \), we write it as \( [(l + m) + (l - m)][(l + m) - (l - m)] \). Simplifying inside the first bracket, \( l + m + l - m \) becomes \( 2l \). Simplifying inside the second bracket, \( l + m - l + m \) becomes \( 2m \). So, the product is \( (2l)(2m) \), which simplifies to \( 4lm \).
In simple words: Treat \( (l+m) \) as one block (A) and \( (l-m) \) as another block (B). Then use the \( A^2 - B^2 \) formula. You get \( (A+B)(A-B) \). When you add \( (l+m) + (l-m) \), you get \( 2l \). When you subtract \( (l+m) - (l-m) \), you get \( 2m \). Multiply them to get \( 4lm \).
6. We are asked to factorise \( 9x^2y^2 - 16 \). This expression can be seen as \( (3xy)^2 - (4)^2 \). This is a difference of squares pattern. Applying the identity \( a^2 - b^2 = (a + b)(a - b) \), we factor it into \( (3xy + 4)(3xy - 4) \).
In simple words: For \( 9x^2y^2 - 16 \), recognize that \( 9x^2y^2 \) is \( (3xy)^2 \) and \( 16 \) is \( (4)^2 \). It's a difference of squares, so it factors into \( (3xy+4)(3xy-4) \).
7. We need to factorise \( (x^2 - 2xy + y^2) - z^2 \). First, notice that the expression \( x^2 - 2xy + y^2 \) is a perfect square trinomial, which is equal to \( (x - y)^2 \). So, the entire expression becomes \( (x - y)^2 - z^2 \). This is now in the form of a difference of squares, \( A^2 - B^2 \), where \( A = (x-y) \) and \( B = z \). Using the identity \( A^2 - B^2 = (A + B)(A - B) \), we factor it into \( [(x - y) + z][(x - y) - z] \). This simplifies to \( (x - y + z)(x - y - z) \).
In simple words: First, change \( x^2 - 2xy + y^2 \) into \( (x-y)^2 \). Then the whole problem becomes \( (x-y)^2 - z^2 \). This is a difference of squares, so it factors into \( (x-y+z)(x-y-z) \).
8. We are given \( 25a^2 - 4b^2 + 28bc - 49c^2 \). First, let's group the terms: \( 25a^2 - (4b^2 - 28bc + 49c^2) \). Now, notice that \( 4b^2 - 28bc + 49c^2 \) is a perfect square trinomial. It can be written as \( (2b)^2 - 2(2b)(7c) + (7c)^2 \), which simplifies to \( (2b - 7c)^2 \). So, the entire expression becomes \( 25a^2 - (2b - 7c)^2 \). This is a difference of squares, where \( A = 5a \) and \( B = (2b - 7c) \). Using \( A^2 - B^2 = (A + B)(A - B) \), we get \( [5a + (2b - 7c)][5a - (2b - 7c)] \). This simplifies to \( (5a + 2b - 7c)(5a - 2b + 7c) \).
In simple words: Group the last three terms and factor out a negative sign: \( 25a^2 - (4b^2 - 28bc + 49c^2) \). The part in the bracket is \( (2b-7c)^2 \). So, the expression becomes \( (5a)^2 - (2b-7c)^2 \). Now, use the difference of squares formula, \( (5a + (2b-7c))(5a - (2b-7c)) \).

Exam Tip: Always look for a greatest common factor (GCF) as the first step in any factorization problem, then check for perfect squares or differences of squares.

Question 3. Factorise the expressions?
1. \( ax^2 + bx \)
2. \( 7p^2 + 21q^2 \)
3. \( 2x^3 + 2xy^2 + 2xz^2 \)
4. \( am^2 + bm^2 + bn^2 + an^2 \)
5. \( (lm + l) + m + l \)
6. \( y(y + z) + 9(y + z) \)
7. \( 5y^2 - 20y - 8z + 2yz \)
8. \( 10ab + 4a + 5b + 2 \)
9. \( 6xy - 4y + 6 - 9x \)
Answer:
1. We need to factorise \( ax^2 + bx \). We observe that both terms have a common factor of \( x \). So, we can take \( x \) out as the common factor. This gives us \( x(ax + b) \).
In simple words: Look for a common part in both \( ax^2 \) and \( bx \). Both have an \( x \). Take \( x \) out, and what's left is \( ax \) from the first term and \( b \) from the second term.
2. We are asked to factorise \( 7p^2 + 21q^2 \). Both terms share a common factor of 7. By factoring out 7, the expression becomes \( 7(p^2 + 3q^2) \).
In simple words: Find the biggest number that divides both 7 and 21, which is 7. Take 7 out of both terms. What remains is \( p^2 \) plus \( 3q^2 \).
3. We need to factorise \( 2x^3 + 2xy^2 + 2xz^2 \). Each term in this expression has \( 2x \) as a common factor. When we factor out \( 2x \), the remaining terms are \( x^2 \), \( y^2 \), and \( z^2 \). So, the factored form is \( 2x(x^2 + y^2 + z^2) \).
In simple words: All three parts have \( 2x \). Take \( 2x \) outside the bracket. Inside, you will have \( x^2 \) (from \( 2x^3 \)), \( y^2 \) (from \( 2xy^2 \)), and \( z^2 \) (from \( 2xz^2 \)).
4. We are asked to factorise \( am^2 + bm^2 + bn^2 + an^2 \). This is a four-term expression, suitable for grouping. First, group the terms: \( (am^2 + bm^2) + (bn^2 + an^2) \). From the first group, we factor out \( m^2 \), giving \( m^2(a + b) \). From the second group, we factor out \( n^2 \), giving \( n^2(b + a) \). Since \( (b+a) \) is the same as \( (a+b) \), we now have \( m^2(a + b) + n^2(a + b) \). We see that \( (a+b) \) is a common binomial factor. Factoring out \( (a+b) \), we obtain \( (a + b)(m^2 + n^2) \).
In simple words: Group the first two terms and the last two terms. From \( am^2 + bm^2 \), take out \( m^2 \). From \( bn^2 + an^2 \), take out \( n^2 \). You will get \( m^2(a+b) + n^2(b+a) \). Since \( b+a \) is the same as \( a+b \), take out \( (a+b) \) as a common factor.
5. Let's factorise \( (lm + l) + (m + l) \). First, consider the term \( lm + l \). We can factor out \( l \) from this, which gives \( l(m + 1) \). So, the original expression becomes \( l(m + 1) + (m + 1) \). Now, we notice that \( (m+1) \) is a common factor in both terms. Factoring out \( (m+1) \), we get \( (m + 1)(l + 1) \).
In simple words: Group the terms \( (lm+l) \) and \( (m+1) \). From \( (lm+l) \), take out \( l \), which leaves \( l(m+1) \). Now you have \( l(m+1) + (m+1) \). Since \( (m+1) \) is common, take it out, and you are left with \( (m+1)(l+1) \).
6. We need to factorise \( y(y + z) + 9(y + z) \). In this expression, we can clearly see that \( (y + z) \) is a common factor in both terms. By factoring out \( (y + z) \), the remaining terms are \( y \) and \( 9 \). So, the factored expression is \( (y + z)(y + 9) \).
In simple words: Notice that \( (y+z) \) is present in both parts. Take it out as a common factor. What remains are \( y \) from the first part and \( 9 \) from the second part.
7. We are asked to factorise \( 5y^2 - 20y - 8z + 2yz \). First, group the terms: \( (5y^2 - 20y) + (-8z + 2yz) \). From the first group, factor out \( 5y \), leaving \( 5y(y - 4) \). From the second group, factor out \( 2z \). This yields \( 2z(-4 + y) \), which is the same as \( 2z(y - 4) \). Now we have \( 5y(y - 4) + 2z(y - 4) \). The common binomial factor is \( (y - 4) \). Factoring it out, we get \( (y - 4)(5y + 2z) \).
In simple words: Group the terms. From \( 5y^2 - 20y \), take out \( 5y \). From \( -8z + 2yz \), take out \( 2z \). You will get \( 5y(y-4) + 2z(y-4) \). Now, \( (y-4) \) is common, so take it out.
8. Let's factorise \( 10ab + 4a + 5b + 2 \). We group the terms: \( (10ab + 4a) + (5b + 2) \). From the first group, we factor out \( 2a \), which gives \( 2a(5b + 2) \). The second group, \( (5b + 2) \), can be written as \( 1(5b + 2) \). Now we have \( 2a(5b + 2) + 1(5b + 2) \). Since \( (5b + 2) \) is a common binomial factor, we factor it out to get \( (5b + 2)(2a + 1) \).
In simple words: Group the first two terms and the last two terms. From \( 10ab + 4a \), take out \( 2a \). You'll get \( 2a(5b+2) \). The other group is just \( (5b+2) \). Now, \( (5b+2) \) is common in both, so take it out, leaving \( (5b+2)(2a+1) \).
9. We are asked to factorise \( 6xy - 4y + 6 - 9x \). First, we rearrange the terms for effective grouping: \( 6xy - 4y - 9x + 6 \). Now, group them: \( (6xy - 4y) + (-9x + 6) \). From the first group, factor out \( 2y \), resulting in \( 2y(3x - 2) \). From the second group, factor out \( -3 \), which yields \( -3(3x - 2) \). Now we have \( 2y(3x - 2) - 3(3x - 2) \). The common binomial factor is \( (3x - 2) \). Factoring it out, we obtain \( (3x - 2)(2y - 3) \).
In simple words: Rearrange the terms to group them like \( (6xy - 4y) \) and \( (-9x + 6) \). From the first group, take out \( 2y \). From the second, take out \( -3 \). You'll see \( (3x-2) \) is common in both. Then, take \( (3x-2) \) out.

Exam Tip: For factorization by grouping, try different combinations of terms if the initial grouping does not yield common binomial factors.

Question 4. Factorise:
1. \( a^4 - b^4 \)
2. \( p^4 - 81 \)
3. \( x^4 - (y + z)^4 \)
4. \( x^4 - (x - z)^4 \)
5. \( a^4 - 2a^2b^2 + b^4 \)
Answer:
1. We need to factorise \( a^4 - b^4 \). This can be written as \( (a^2)^2 - (b^2)^2 \), which is a difference of squares. Using the identity \( a^2 - b^2 = (a - b)(a + b) \), we get \( (a^2 + b^2)(a^2 - b^2) \). Now, the term \( (a^2 - b^2) \) is itself a difference of squares. Factoring it further, we get \( (a + b)(a - b) \). So, the complete factorization is \( (a^2 + b^2)(a + b)(a - b) \).
In simple words: First, treat \( a^4 - b^4 \) as \( (a^2)^2 - (b^2)^2 \). Use the difference of squares formula to get \( (a^2+b^2)(a^2-b^2) \). Then, notice that \( (a^2-b^2) \) can be factored again into \( (a-b)(a+b) \).
2. We are asked to factorise \( p^4 - 81 \). This expression can be written as \( (p^2)^2 - (9)^2 \), which is a difference of squares. Applying the identity \( a^2 - b^2 = (a + b)(a - b) \), we get \( (p^2 + 9)(p^2 - 9) \). Now, the factor \( (p^2 - 9) \) is also a difference of squares. We factor it as \( (p)^2 - (3)^2 \), which becomes \( (p + 3)(p - 3) \). So, the full factorization is \( (p^2 + 9)(p + 3)(p - 3) \).
In simple words: Change \( p^4 - 81 \) to \( (p^2)^2 - (9)^2 \). Use the difference of squares formula to get \( (p^2+9)(p^2-9) \). Then, factor \( (p^2-9) \) again into \( (p-3)(p+3) \).
3. We need to factorise \( x^4 - (y + z)^4 \). This can be written as \( (x^2)^2 - [(y + z)^2]^2 \), which is a difference of squares. Applying \( A^2 - B^2 = (A + B)(A - B) \), where \( A = x^2 \) and \( B = (y + z)^2 \), we get \( [x^2 + (y + z)^2][x^2 - (y + z)^2] \). The second factor, \( [x^2 - (y + z)^2] \), is again a difference of squares, with \( A = x \) and \( B = (y + z) \). Factoring this, we get \( [x + (y + z)][x - (y + z)] \), which simplifies to \( (x + y + z)(x - y - z) \). Therefore, the full factorization is \( [x^2 + (y + z)^2](x + y + z)(x - y - z) \).
In simple words: Treat \( x^4 - (y+z)^4 \) as a difference of squares: \( (x^2)^2 - ((y+z)^2)^2 \). This gives \( (x^2+(y+z)^2)(x^2-(y+z)^2) \). The second part, \( (x^2-(y+z)^2) \), can be factored again as \( (x+(y+z))(x-(y+z)) \).
4. We need to factorise \( x^4 - (x - z)^4 \). This can be written as \( (x^2)^2 - [(x - z)^2]^2 \), which is a difference of squares. Applying \( A^2 - B^2 = (A + B)(A - B) \), where \( A = x^2 \) and \( B = (x - z)^2 \), we get \( [x^2 + (x - z)^2][x^2 - (x - z)^2] \). Now, let's factorise the second term, \( x^2 - (x - z)^2 \). This is a difference of squares with \( A = x \) and \( B = (x - z) \). So it becomes \( [x + (x - z)][x - (x - z)] \). Simplifying inside the brackets: \( (x + x - z) \) becomes \( (2x - z) \), and \( (x - x + z) \) becomes \( z \). Thus, \( x^2 - (x - z)^2 = z(2x - z) \). Combining all factors, the complete factorization is \( [x^2 + (x - z)^2]z(2x - z) \).
In simple words: First, treat \( x^4 - (x-z)^4 \) as a difference of squares: \( (x^2)^2 - ((x-z)^2)^2 \). This results in \( (x^2+(x-z)^2)(x^2-(x-z)^2) \). Then, factor the second part, \( (x^2-(x-z)^2) \), into \( (x+(x-z))(x-(x-z)) \), which simplifies to \( z(2x-z) \).
5. We are given the expression \( a^4 - 2a^2b^2 + b^4 \). This can be identified as a perfect square trinomial. It can be written as \( (a^2)^2 - 2(a^2)(b^2) + (b^2)^2 \). This precisely matches the algebraic identity \( (A - B)^2 = A^2 - 2AB + B^2 \), where \( A = a^2 \) and \( B = b^2 \). Therefore, the expression simplifies to \( (a^2 - b^2)^2 \). Further, \( (a^2 - b^2) \) is a difference of squares, which factors into \( (a - b)(a + b) \). So, \( (a^2 - b^2)^2 \) becomes \( [(a - b)(a + b)]^2 \), which can also be written as \( (a - b)^2(a + b)^2 \).
In simple words: Recognize \( a^4 - 2a^2b^2 + b^4 \) as a perfect square of \( (a^2 - b^2) \). So it's \( (a^2-b^2)^2 \). Since \( a^2-b^2 \) factors into \( (a-b)(a+b) \), the final factorization is \( ((a-b)(a+b))^2 \).

Exam Tip: Be very careful with negative signs when distributing them, especially in the difference of squares formula, like in \( x - (x-z) \).

Question 5. Factorise the following expressions.
1. \( p^2 + 6p + 8 \)
2. \( q^2 - 10q + 21 \)
3. \( p^2 + 6p - 16 \)
Answer:
1. We are asked to factorise \( p^2 + 6p + 8 \). We can rewrite the constant term 8 as \( 9 - 1 \) to complete the square for the first two terms: \( p^2 + 6p + 9 - 1 \). The first three terms \( p^2 + 6p + 9 \) form a perfect square trinomial, \( (p + 3)^2 \). So the expression becomes \( (p + 3)^2 - 1^2 \). This is a difference of squares. Using \( A^2 - B^2 = (A + B)(A - B) \), where \( A = (p + 3) \) and \( B = 1 \), we get \( [(p + 3) + 1][(p + 3) - 1] \). This simplifies to \( (p + 4)(p + 2) \).
In simple words: For \( p^2 + 6p + 8 \), find two numbers that multiply to 8 and add up to 6. These numbers are 4 and 2. So, it factors into \( (p+4)(p+2) \).
2. We need to factorise \( q^2 - 10q + 21 \). We can rewrite 21 as \( 25 - 4 \) to complete the square: \( q^2 - 10q + 25 - 4 \). The first three terms \( q^2 - 10q + 25 \) form a perfect square trinomial, which is \( (q - 5)^2 \). So the expression becomes \( (q - 5)^2 - 2^2 \). This is a difference of squares. Using \( A^2 - B^2 = (A + B)(A - B) \), where \( A = (q - 5) \) and \( B = 2 \), we get \( [(q - 5) + 2][(q - 5) - 2] \). This simplifies to \( (q - 3)(q - 7) \).
In simple words: For \( q^2 - 10q + 21 \), find two numbers that multiply to 21 and add up to -10. These numbers are -3 and -7. So, it factors into \( (q-3)(q-7) \).
3. Let's factorise \( p^2 + 6p - 16 \). We can rewrite -16 as \( 9 - 25 \) to complete the square: \( p^2 + 6p + 9 - 25 \). The first three terms \( p^2 + 6p + 9 \) form a perfect square trinomial, \( (p + 3)^2 \). So the expression becomes \( (p + 3)^2 - 5^2 \). This is a difference of squares. Using \( A^2 - B^2 = (A + B)(A - B) \), where \( A = (p + 3) \) and \( B = 5 \), we get \( [(p + 3) + 5][(p + 3) - 5] \). This simplifies to \( (p + 8)(p - 2) \).
In simple words: For \( p^2 + 6p - 16 \), find two numbers that multiply to -16 and add up to 6. These numbers are 8 and -2. So, it factors into \( (p+8)(p-2) \).

Exam Tip: For quadratic trinomials \( x^2 + bx + c \), always look for two numbers that multiply to \( c \) and add up to \( b \). Remember to consider the signs carefully.