EXERCISE 12.2 — Solutions

1. Factorise the following expressions.

(i) \(a^{2} + 8a + 16\)

Recognise a perfect square: \(a^{2} + 2\times a \times 4 + 4^{2}\).

Rearrange: \(a^{2} + 4^{2} + 2\times a \times 4\).

Using identity \(a^{2} + b^{2} + 2ab = (a + b)^{2}\) with \(a = a\) and \(b = 4\).

So it is \((a + 4)^{2}\).

Final Answer: \(a^{2} + 8a + 16 = (a + 4)^{2}.\)

(ii) \(p^{2} - 10p + 25\)

Recognise a perfect square: \(p^{2} - 2\times p \times 5 + 5^{2}\).

Rearrange: \(p^{2} + 5^{2} - 2\times p \times 5\).

Using identity \(a^{2} + b^{2} - 2ab = (a - b)^{2}\) with \(a = p\) and \(b = 5\).

So it is \((p - 5)^{2}\).

Final Answer: \(p^{2} - 10p + 25 = (p - 5)^{2}.\)

(iii) \(25m^{2} + 30m + 9\)

Recognise a perfect square: \((5m)^{2} + 2\times(5m)\times 3 + 3^{2}\).

Rearrange: \((5m)^{2} + 3^{2} + 2\times(5m)\times 3\).

Using identity \(a^{2} + b^{2} + 2ab = (a + b)^{2}\) with \(a = 5m\) and \(b = 3\).

So it is \((5m + 3)^{2}\).

Final Answer: \(25m^{2} + 30m + 9 = (5m + 3)^{2}.\)

(iv) \(49y^{2} + 84yz + 36z^{2}\)

Recognise a perfect square: \((7y)^{2} + 2\times(7y)\times(6z) + (6z)^{2}\).

Rearrange: \((7y)^{2} + (6z)^{2} + 2\times(7y)\times(6z)\).

Using identity \(a^{2} + b^{2} + 2ab = (a + b)^{2}\) with \(a = 7y\) and \(b = 6z\).

So it is \((7y + 6z)^{2}\).

Final Answer: \(49y^{2} + 84yz + 36z^{2} = (7y + 6z)^{2}.\)

(v) \(4x^{2} - 8x + 4\)

Factor out common 4 first: \(4(x^{2} - 2x + 1)\).

Rearrange inside: \(x^{2} + 1^{2} - 2\times x \times 1\).

Using identity \(a^{2} + b^{2} - 2ab = (a - b)^{2}\) with \(a = x, b = 1\).

So it is \(4(x - 1)^{2}\).

Final Answer: \(4x^{2} - 8x + 4 = 4(x - 1)^{2}.\)

(vi) \(121b^{2} - 88bc + 16c^{2}\)

Recognise a perfect square: \((11b)^{2} - 2\times(11b)\times(4c) + (4c)^{2}\).

Rearrange: \((11b)^{2} + (4c)^{2} - 2\times(11b)\times(4c)\).

Using identity \(a^{2} + b^{2} - 2ab = (a - b)^{2}\) with \(a = 11b, b = 4c\).

So it is \((11b - 4c)^{2}\).

Final Answer: \(121b^{2} - 88bc + 16c^{2} = (11b - 4c)^{2}.\)

(vii) \((l + m)^{2} - 4lm\) (Hint: expand first)

Expand first: \((l + m)^{2} = l^{2} + 2lm + m^{2}\).

Subtract \(4lm\): \(l^{2} + 2lm + m^{2} - 4lm = l^{2} - 2lm + m^{2}\).

Using identity \(a^{2} + b^{2} - 2ab = (a - b)^{2}\) with \(a = l, b = m\).

So it is \((l - m)^{2}\).

Final Answer: \((l + m)^{2} - 4lm = (l - m)^{2}.\)

(viii) \(a^{4} + 2a^{2}b^{2} + b^{4}\)

See as squares of squares: \((a^{2})^{2} + 2\times a^{2}\times b^{2} + (b^{2})^{2}\).

Rearrange: \((a^{2})^{2} + (b^{2})^{2} + 2\times a^{2}\times b^{2}\).

Using identity \(A^{2} + B^{2} + 2AB = (A + B)^{2}\) with \(A = a^{2}, B = b^{2}\).

So it is \((a^{2} + b^{2})^{2}\).

Final Answer: \(a^{4} + 2a^{2}b^{2} + b^{4} = (a^{2} + b^{2})^{2}.\)

2. Factorise.

(i) \(4p^{2} - 9q^{2}\)

Recognise & rewrite squares: \((2p)^{2} - (3q)^{2}\).

Using Identity: \(a^{2} - b^{2} = (a - b)(a + b)\) with \(a = 2p,\; b = 3q\).

Apply identity: \((2p - 3q)(2p + 3q)\).

Final Answer: \(4p^{2} - 9q^{2} = (2p - 3q)(2p + 3q).\)

(ii) \(63a^{2} - 112b^{2}\)

Take out common factor: \(7(9a^{2} - 16b^{2})\).

Using Identity: \(a^{2} - b^{2} = (a - b)(a + b)\) with \(a = 3a,\; b = 4b\).

Apply identity to bracket: \((3a - 4b)(3a + 4b)\).

Final Answer: \(63a^{2} - 112b^{2} = 7(3a - 4b)(3a + 4b).\)

(iii) \(49x^{2} - 36\)

Rewrite as squares: \((7x)^{2} - 6^{2}\).

Using Identity: \(a^{2} - b^{2} = (a - b)(a + b)\) with \(a = 7x,\; b = 6\).

Apply identity: \((7x - 6)(7x + 6)\).

Final Answer: \(49x^{2} - 36 = (7x - 6)(7x + 6).\)

(iv) \(16x^{5} - 144x^{3}\)

Take out common factor: \(16x^{3}(x^{2} - 9)\).

Using Identity: \(a^{2} - b^{2} = (a - b)(a + b)\) with \(a = x,\; b = 3\).

Apply identity: \(x^{2} - 9 = (x - 3)(x + 3)\).

Final Answer: \(16x^{5} - 144x^{3} = 16x^{3}(x - 3)(x + 3).\)

(v) \((l + m)^{2} - (l - m)^{2}\)

Recognise difference of squares with \(A = l + m,\; B = l - m\).

Using Identity: \(A^{2} - B^{2} = (A - B)(A + B)\).

Apply identity: \( [(l + m) - (l - m) ] [ (l + m) + (l - m) ] \)

Simplify each bracket:

First bracket: \((l + m) - (l - m) = l + m - l + m = 2m\).

Second bracket: \((l + m) + (l - m) = l + m + l - m = 2l\).

Now we can write : \(A - B = 2m,\; A + B = 2l\). So expression \(= 2m \times 2l = 4lm\).

Final Answer: \((l + m)^{2} - (l - m)^{2} = 4lm.\)

(vi) \(9x^{2}y^{2} - 16\)

Rewrite as squares: \((3xy)^{2} - 4^{2}\).

Using Identity: \(a^{2} - b^{2} = (a - b)(a + b)\) with \(a = 3xy,\; b = 4\).

Apply identity: \((3xy - 4)(3xy + 4)\).

Final Answer: \(9x^{2}y^{2} - 16 = (3xy - 4)(3xy + 4).\)

(vii) \((x^{2} - 2xy + y^{2}) - z^{2}\)

First note: \(x^{2} - 2xy + y^{2} = (x - y)^{2}\).

Using Identity: \(A^{2} - B^{2} = (A - B)(A + B)\) with \(A = x - y,\; B = z\).

Apply identity: \((x - y - z)(x - y + z)\).

Final Answer: \((x^{2} - 2xy + y^{2}) - z^{2} = (x - y - z)(x - y + z).\)

(viii) \(25a^{2} - 4b^{2} + 28bc - 49c^{2}\)

Rearrange: \(25a^{2} - (4b^{2} - 28bc + 49c^{2})\).

Rewrite: \(4b^{2} - 28bc + 49c^{2} \) as \( {(2b)}^2 - 2 \times 2b \times ({7c})^2 \) .

Recognise identity for this: \(a^{2} - 2ab + b^{2} = (a - b)^{2}\) with \(a = 2b,\; b = 7c\).

So we get: \((2b - 7c)^{2}\).

Now we can rewrite the question as: \(25a^{2} - (2b - 7c)^{2}\).

Again using Identity: \(A^{2} - B^{2} = (A - B)(A + B)\) with \(A = 5a,\; B = (2b - 7c)\).

So we have \(25a^{2} - (2b - 7c)^{2}\).

Apply identity: \((5a - (2b - 7c))(5a + (2b - 7c))\).

Simplify: \((5a - 2b + 7c)(5a + 2b - 7c)\).

Final Answer: \(25a^{2} - 4b^{2} + 28bc - 49c^{2} = (5a - 2b + 7c)(5a + 2b - 7c).\)

3. Factorise the expressions.

(i) \(ax^{2} + bx\)

Common factor is: \(x\).

Take out \(x\): \(x(ax + b)\).

Final Answer: \(ax^{2} + bx = x(ax + b).\)

(ii) \(7p^{2} + 21q^{2}\)

Common factor is: \(7\).

Take out \(7\): \(7(p^{2} + 3q^{2})\).

Final Answer: \(7p^{2} + 21q^{2} = 7(p^{2} + 3q^{2}).\)

(iii) \(2x^{3} + 2xy^{2} + 2xz^{2}\)

Common factor is: \(2x\).

Take out \(2x\): \(2x(x^{2} + y^{2} + z^{2})\).

Final Answer: \(2x^{3} + 2xy^{2} + 2xz^{2} = 2x(x^{2} + y^{2} + z^{2}).\)

(iv) \(am^{2} + bm^{2} + bn^{2} + an^{2}\)

Group terms: \((am^{2} + bm^{2}) + (bn^{2} + an^{2})\).

Factor each: \(m^{2}(a+b) + n^{2}(a+b)\).

Take out common \((a+b)\): \((a+b)(m^{2} + n^{2})\).

Final Answer: \(am^{2} + bm^{2} + bn^{2} + an^{2} = (a+b)(m^{2}+n^{2}).\)

(v) \((lm + l) + m + 1\)

Group terms: \((lm + l) + (m + 1)\).

Factor each: \(l(m+1) + 1(m+1)\).

Take out common \((m+1)\): \((m+1)(l+1)\).

Final Answer: \((lm + l) + m + 1 = (m+1)(l+1).\)

(vi) \(y(y + z) + 9(y + z)\)

Common factor is: \((y+z)\).

Take out \((y+z)\): \((y+z)(y+9)\).

Final Answer: \(y(y+z) + 9(y+z) = (y+z)(y+9).\)

(vii) \(5y^{2} - 20y - 8z + 2yz\)

Group terms: \((5y^{2} - 20y) + (2yz - 8z)\).

Factor each: \(5y(y-4) + 2z(y-4)\).

Take out common \((y-4)\): \((y-4)(5y+2z)\).

Final Answer: \(5y^{2} - 20y - 8z + 2yz = (y-4)(5y+2z).\)

(viii) \(10ab + 4a + 5b + 2\)

Group terms: \((10ab + 4a) + (5b + 2)\).

Factor each: \(2a(5b+2) + 1(5b+2)\).

Take out common \((5b+2)\): \((5b+2)(2a+1)\).

Final Answer: \(10ab + 4a + 5b + 2 = (5b+2)(2a+1).\)

(ix) \(6xy - 4y + 6 - 9x\)

Group terms: \((6xy - 4y) + (6 - 9x)\).

Factor each: \(2y(3x-2) + 3(2-3x)\).

Note: \(3(2-3x) = -3(3x-2)\).

So expression = \((3x-2)(2y-3)\).

Final Answer: \(6xy - 4y + 6 - 9x = (3x-2)(2y-3).\)

4. Factorise.

(i) \(a^{4} - b^{4}\)

Recognise & rewrite squares: \(a^{4} - b^{4} = (a^{2})^{2} - (b^{2})^{2}\).

Using Formula: \(A^{2} - B^{2} = (A - B)(A + B)\) with \(A = a^{2}, B = b^{2}\).

So it becomes: \((a^{2} + b^{2})(a^{2} - b^{2})\).

Again apply formula on \((a^{2} - b^{2})\): \((a - b)(a + b)\).

Final Answer: \(a^{4} - b^{4} = (a - b)(a + b)(a^{2} + b^{2}).\)

(ii) \(p^{4} - 81\)

Recognise: \(81 = 9^{2}\). Write as \(p^{4} - 9^{2} = (p^{2})^{2} - 9^{2}\).

Using Formula: \(a^{2} - b^{2} = (a - b)(a + b)\) with \(a = p^{2}, b = 9\).

So it becomes: \((p^{2} - 9)(p^{2} + 9)\).

Again apply formula to \((p^{2} - 9)\) as \((p^{2} - 3^2 )\) = \((p - 3)(p + 3)\).

Final Answer: \(p^{4} - 81 = (p - 3)(p + 3)(p^{2} + 9).\)

(iii) \(x^{4} - (y+z)^{4}\)

Recognise & rewrite: \((x^{2})^{2} - \big((y+z)^{2}\big)^{2}\).

Using Formula: \(a^{2} - b^{2} = (a - b)(a + b)\) with \(a = x^{2}, b = (y+z)^{2}\).

So factors: \((x^{2} - (y+z)^{2})(x^{2} + (y+z)^{2})\).

Further factor: \(x^{2} - (y+z)^{2} = (x - (y+z))(x + (y+z)) = (x - y - z)(x + y + z)\).

Final Answer: \(x^{4} - (y+z)^{4} = (x - y - z)(x + y + z)\big(x^{2} + (y+z)^{2}\big).\)

(iv) \(x^{4} - (x - z)^{4}\)

Rewrite: \((x^{2})^{2} - \big((x - z)^{2}\big)^{2}\).

Using Formula: \(a^{2} - b^{2} = (a - b)(a + b)\) with \(a = x^{2}, b = (x - z)^{2}\).

So factors: \((x^{2} - (x - z)^{2})(x^{2} + (x - z)^{2})\).

Simplify first: \(x^{2} - (x^{2} - 2xz + z^{2}) = 2xz - z^{2} = z(2x - z)\).

Final Answer: \(x^{4} - (x - z)^{4} = z(2x - z)\big(x^{2} + (x - z)^{2}\big).\)

(v) \(a^{4} - 2a^{2}b^{2} + b^{4}\)

Recognise: \(a^{4} - 2a^{2}b^{2} + b^{4} = (a^{2})^{2} - 2(a^{2})(b^{2}) + (b^{2})^{2}\).

This is a perfect square trinomial.

Using Identity: \(A^{2} - 2AB + B^{2} = (A - B)^{2}\) with \(A = a^{2}, B = b^{2}\).

So it becomes: \((a^{2} - b^{2})^{2}\).

Again apply difference of squares: \(a^{2} - b^{2} = (a - b)(a + b)\).

Final Answer: \(a^{4} - 2a^{2}b^{2} + b^{4} = (a - b)^{2}(a + b)^{2}.\)

5. Factorise the following expressions.

(i) \(p^{2} + 6p + 8\)

Split the middle term such that product is \(8\) and sum is \(6\) → numbers are \(2\) and \(4\).

Check: \(2 \times 4 = 8,\; 2 + 4 = 6.\)

So rewrite as: \(p^{2} + 2p + 4p + 8\).

Group: \((p^{2} + 2p) + (4p + 8)\).

Take common factors: \(p(p + 2) + 4(p + 2)\).

So we have: \((p + 2)(p + 4)\).

Final Answer: \(p^{2} + 6p + 8 = (p + 2)(p + 4).\)

(ii) \(q^{2} - 10q + 21\)

Split the middle term such that product is \(21\) and sum is \(-10\) → numbers are \(-3\) and \(-7\).

Check: \((-3) \times (-7) = 21,\; -3 + -7 = -10.\)

So rewrite as: \(q^{2} - 3q - 7q + 21\).

Group: \((q^{2} - 3q) + (-7q + 21)\).

Take common factors: \(q(q - 3) - 7(q - 3)\).

So we have: \((q - 3)(q - 7)\).

Final Answer: \(q^{2} - 10q + 21 = (q - 3)(q - 7).\)

(iii) \(p^{2} + 6p - 16\)

Split the middle term such that product is \(-16\) and sum is \(6\) → numbers are \(8\) and \(-2\).

Check: \(8 \times -2 = -16,\; 8 + (-2) = 6.\)

So rewrite as: \(p^{2} + 8p - 2p - 16\).

Group: \((p^{2} + 8p) + (-2p - 16)\).

Take common factors: \(p(p + 8) - 2(p + 8)\).

So we have: \((p + 8)(p - 2)\).

Final Answer: \(p^{2} + 6p - 16 = (p + 8)(p - 2).\)

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