Question 512 Marks
Expand:
$\Big(1+\frac{2}{3}{\text{a}}\Big)^3$
Answer$\Big(1+\frac{2}{3}{\text{a}}\Big)^3$ $=\Big(\frac{2}{3}\text{a}\Big)^3+3\times\Big(\frac{2}{3}\text{a}\Big)^2\times1+3\text{a}\frac{2}{3}\text{a}\times(1)^2+(1)^3$ $=\frac{8}{27}\text{a}^3+\frac{4}{3}\text{a}^2+2\text{a}+1$
View full question & answer→Question 522 Marks
Factorise:$3a^3b - 243ab^3$
Answer$3a^3b - 243ab^3$
$= 3ab(a^2 - 81b^2)$
$= 3ab[(a)^2 - (9b)^2]$
$= 3ab(a + 9b)(a - 9b)$ $\big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
View full question & answer→Question 532 Marks
If $a, b, c$ are all nonzero and $a + b + c = 0$, prove that:
$\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3$
Answer$a + b + c = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$
Thus,
We have:
$\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}}$
$=3$
View full question & answer→Question 542 Marks
Factorise:
$2\text{x}^2+3\sqrt{3}\text{x}+3$
Answer$2\text{x}^2+3\sqrt{3}\text{x}+3$
$2\text{x}^2+2\sqrt{3}\text{x}+\sqrt{3}\text{x}+3$
$=2\text{x}\big(\text{x}+\sqrt{3}\big)+\sqrt{3}\big(\text{x}+\sqrt{3}\big)$
$=\big(\text{x}+\sqrt{3}\big)\big(2\text{x}+\sqrt{3}\big)$
View full question & answer→Question 552 Marks
Evaluate:$(107)^2$
Answer$(107)^2 = (100 + 7)^2$
$= (100)^2 + 2 \times (100) \times (7) + (7)^2$
$= 10000 + 1400 + 49$
$= 11449$
View full question & answer→Question 562 Marks
Factorise:$x^2 - 32x - 105$
Answer$x^2 - 32x - 105$
$= x^2 - 35x + 3x - 105$
$= x(x - 35) + 3(x - 35)$
$= x(x - 35)(x + 3)$
View full question & answer→Question 572 Marks
Factorise:
$6\sqrt{3}\text{x}^2-47\text{x}+5\sqrt{3}$
Answer$6\sqrt{3}\text{x}^2-47\text{x}+5\sqrt{3}$
$=6\sqrt{3}\text{x}^2-45\text{x}-2\text{x}+5\sqrt{3}$
$=3\sqrt{3}\text{x}\big(2\text{x}-5\sqrt{3}\big)-1\big(2\text{x}-5\sqrt{3}\big)$
$=\big(2\text{x}-5\sqrt{3}\big)\big(3\sqrt{3}\text{x}-1\big)$
View full question & answer→Question 582 Marks
Factorise:$(a + b)^3 - a - b$
Answer$(a + b)^3 - a - b$
$= (a + b)^3 - (a + b)$
$= (a + b)[(a + b)^2 - 1^2] $$\big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
$= (a + b)(a + b + 1)(a + b - 1)$
View full question & answer→Question 592 Marks
Factorise:
$2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
Answer$x =2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
$=\big(\sqrt{2}\text{a}\big)^3+\big(2\sqrt{2}\text{b}\big)^3+\text{c}^3-3\times\big(\sqrt{2}\text{a}\big)\times\big(2\sqrt{2}\text{b}\big)\times(\text{c})$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\Big[\big(\sqrt{2}\text{a}\big)^2+(2\sqrt{2}\text{b})^2+\text{c}^2\\-\big(\sqrt{2}\text{a}\big)\times\big(2\sqrt{2}\text{b}\big)-\big(2\sqrt{2}\text{b}\big)\times(\text{c})-\big(\sqrt{2}\text{a}\big)\times(\text{c})\Big]$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2\text{bc}}-\sqrt{2}\text{ac}\big)$
View full question & answer→Question 602 Marks
Factorise:$x^2 - x - 156$
Answer$x^2 - x - 156$
$= x^2 - 13x + 12x - 156$
$= x(x - 13) + 12(x - 13)$
$= (x - 13)(x + 12)$
View full question & answer→Question 612 Marks
Factorise:
$\text{x}^2+\sqrt{2}\text{x}-24$
Answer$\text{x}^2+\sqrt{2}\text{x}-24$
$=\text{x}^2+4\sqrt{2}\text{x}-3\sqrt{2}\text{x}-24$
$=\text{x}(\text{x}+4\sqrt{2})-3\sqrt{2}(\text{x}+4\sqrt{2})$
$=(\text{x}-4\sqrt{2})(\text{x}-3\sqrt{2})$
View full question & answer→Question 622 Marks
Factorise:
$\frac{3}{2}\text{x}^2+16\text{x}+10$
Answer$\frac{3}{2}\text{x}^2+16\text{x}+10$
$=\frac{3}{2}\text{x}^2+\text{x}+15\text{x}+10$
$=\frac{\text{x}}{2}(3\text{x}+2)+5(3\text{x}+2)$
$(3\text{x}+2)\Big(\frac{\text{x}}{2}+5\Big)$
View full question & answer→Question 632 Marks
Factorise:$x - 8xy^3$
Answer$x - 8xy^3$
$= x(1 - 8y^3)$
$= x[(1)^3 - (2y)^3]$
$= x(1 - 2y)[(1)^2 + 1 \times 2y + (2y)^2]$ Since $a^3 - b^3 = (a - b)(a^2 + a \times b + b^2)$
$= x (1 - 2y)(1 + 2y + 4y^2)$
View full question & answer→Question 642 Marks
Factorise:
$\text{x}^4+\frac{4}{\text{x}^4}$
Answer$\text{x}^4+\frac{4}{\text{x}^4}$
$=\text{x}^4+\frac{4}{\text{x}^4}+4-4$
$=\big(\text{x}^2\big)^2+\Big(\frac{2}{\text{x}^2}\Big)^2+2\big(\text{x}^2\big)\Big(\frac{2}{\text{x}^2}\Big)-2^2$
$=\bigg[\big(\text{x}^2\big)^2+\Big(\frac{2}{\text{x}^2}\Big)^2+2\big(\text{x}^2\big)\Big(\frac{2}{\text{x}^2}\Big)\bigg]-2^2$
$=\Big[\text{x}^2+\frac{2}{\text{x}^2}\Big]^2-2^2$
$=\Big(\text{x}^2+\frac{2}{\text{x}^2}+2\Big)\Big(\text{x}^2+\frac{2}{\text{x}^2}-2\Big)$
View full question & answer→Question 652 Marks
Factorise: $x^6 - 729$
Answer$x^6 - 729$
$= (x^2)^3 - (9)^3$
$= (x^2 - 9)[(x^2)^2 + x^2 \times 9 + (9)^2] $ Since $ a^3 - b^3 = (a - b)(a^2 + a \times b + b^2)$
$= (x^2 - 9)(x^4 + 9x^2 + 81)$
$= (x + 3)(x - 3)[(x^2 + 9)^2 - (3x)^2]$
$= (x + 3)(x - 3)(x^2 + 3x + 9)(x^2 - 3x + 9)$
View full question & answer→Question 662 Marks
Factorise:$27a^3 - b^3 + 8c^3 - 18abc$
Answer$27a^3 - b^3 + 8c^3 - 18abc$
$= (3a)^3 + (-b)^3 + (2c)^3 - 3 \times (3a) \times (-b) \times (2c)$
$= [3a + (-b) + 2c][(3a)^2 + (-b)^2 + (2c)^2 - 3a(-b)2c - 3a \times 2c]$
$= (3a - b + 2c)(9a^2 + b^2 + 4c^2 + 3ab + 2bc - 6ac)$
View full question & answer→Question 672 Marks
Factorise:
$1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$
Answer$1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$
$=(1)^3+\Big(\frac{3}{5}\text{a}\Big)^3+3(1)^2\Big(\frac{3}{5}\text{a}\Big)+3(1)\Big(\frac{3}{5}\text{a}\Big)^2$
$=\Big(1+\frac{3}{5}\text{a}\Big)^3$
Hence, factorisation of $1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$ is $=\Big(1+\frac{3}{5}\text{a}\Big)^3$
View full question & answer→Question 682 Marks
Factorise:
$21\text{x}^2-2\text{x}+\frac{1}{21}$
Answer$21\text{x}^2-2\text{x}+\frac{1}{21}$
$21\text{x}^2-\text{x}-\text{x}+\frac{1}{21}$
$=21\text{x}\Big(\text{x}-\frac{1}{21}\Big)-1\Big(\text{x}-\frac{1}{21}\Big)$
$=\Big(\text{x}-\frac{1}{21}\Big)(21\text{x}-1)$
View full question & answer→Question 692 Marks
Factorise:$a^3 + 0.008$
Answer$a^3 + 0.008$
$= (a)^3 + (0.2)^3$
$= (a + 0.2)[(a)^2 - a \times 0.2 + (0.2)^2] $Since $a^3 + b^3 = (a + b)(a^2 - a \times b + b^2)$
$= (a + 0.2)(a^2 - 0.2a + 0.04)$
View full question & answer→Question 702 Marks
Expand:$(2b - b + c)^2$
Answer$(2b - b + c)^2 = [(2a) + (-b) + (c)]^2$
$= (2a)^2 + (-b)^2 + (c)^2 + 2(2a)(-b) + 2(-b)(c) + 4(a)(c)$
$= 14a^2 + b^2 + c^2 - 4ab - 2bc + 4ac$
View full question & answer→Question 712 Marks
Factorise:
$\sqrt{3}\text{x}^2-10\text{x}+8\sqrt{3}$
Answer$\sqrt{3}\text{x}^2-10\text{x}+8\sqrt{3}$
$=\sqrt{3}\text{x}^2+4\text{x}+6\text{x}+8\sqrt{3}$
$=\text{x}\big(\sqrt{3}\text{x}+4\big)+2\sqrt{3}\big(\sqrt{3}\text{x}+4\big)$
$=\big(\sqrt{3}\text{x}+4\big)\big(\text{x}+2\sqrt{3}\big)$
View full question & answer→Question 722 Marks
Factorise:
$\text{x}^2+5\sqrt{5}\text{x}+30$
Answer$\text{x}^2+5\sqrt{5}\text{x}+30$
$=\text{x}^2+2\sqrt{5}\text{x}+3\sqrt{5}\text{x}+30$
$=\text{x}(\text{x}+2\sqrt{5})+3\sqrt{5}(\text{x}+2\sqrt{5})$
$=(\text{x}+2\sqrt{5})(\text{x}+3\sqrt{5})$
View full question & answer→Question 732 Marks
Factorise:$125 - 8x^3 - 27y^3 - 90xy$
Answer$125 - 8x^3 - 27y^3 - 90xy$
$= 5^3 + (-2x)^3 + (-3y)^3 - 3 \times 5 \times (-2x) \times (-3y)$
$= [5 + (-2x) + (-3y)][5^2 + (-2x)^2 + (-3y)^2 - 5 \times (-2x) - (-2x)(-3y) - 5 \times (-3y)]$
$= (5 - 2x - 3y)(25 + 4x^2 + 9y^2 + 10x - 6xy + 15y)$
View full question & answer→Question 742 Marks
Factorise:$25x^2 - 10x + 1 - 36y^2$
Answer$25x^2 - 10x + 1 - 36y^2$
$= (25x^2 - 10x + 1) - 36y^2$
$= [(5x)^2 - 2(5x)(1) + (1)^2] - (6y)^2$
$= (5x - 1)^2 - (6y)^2$
$= (5x - 1 - 6y)(5x - 1 + 6y)$
View full question & answer→Question 752 Marks
Factorise:$4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$
AnswerWe have:
$4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$
$= (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)$
$= [(2x) + (3y) + (-4z)]^2$
$= (2x + 3y - 4z)^2$
View full question & answer→Question 762 Marks
Find the product.$(x + y - z)(x^2 + y^2 + z^2 - xy + yz + zx)$
Answer$(x + y - z)(x^2 + y^2 + z^2 - xy + yz + zx)$
$= [x + y + (-z)][x^2 + y^2 + (-z)^2 - xy - y \times (-z) - [-z] \times x]$
$= x^3 + y^3 + (-z)^3 - 3x \times y \times (-z)$
$= x^3 + y^3 + (-z)^3 - 3x \times y \times (-z)$
$= x^3 + y^3 - z^3 + 3xyz$
View full question & answer→Question 772 Marks
Factorise: $125\text{a}^3+\frac{1}{8}$
Answer$125\text{a}^3+\frac{1}{8}$
We know that:
Since $a^2 + b^3 = (a + b)(a^2 - a \times b + b^2)$
Let us rewrite
$125\text{a}^3+\frac{1}{8}$
$=(5\text{a})^3+\Big(\frac{1}{2}\Big)^3$
$=\Big(5\text{a}+\frac{1}{2}\Big)\bigg[(5\text{a})^2-5\text{a}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\bigg]$
$=\Big(5\text{a}+\frac{1}{2}\Big)\Big(25\text{a}^2-\frac{5\text{a}}{2}+\frac{1}{4}\Big)$
View full question & answer→Question 782 Marks
Factorise:$x^4 - 625$
Answer$x^4 - 625$
$= (x^2)^2 - (25)^2$
$= (x^2 + 25)(x^2 - 25)$
$= (x^2 + 25)(x^2 - 5^2)$ $\Big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\Big]$
$= (x^2 + 25)(x + 5)(x - 5)$
View full question & answer→Question 792 Marks
Factorise:$9a^2 + 3a - 8b - 64b^2$
Answer$9a^2 + 3a - 8b - 64b^2$
$= 9a^2 - 64b^2 + 3a - 8b$
$= (3a)^2 - (8b)^2 + (3a - 8b) \big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
$= (3a + 8b)(3a - 8b) + (3a - 8b)$
$= (3a - 8b)(3a + 8b + 1)$
View full question & answer→Question 802 Marks
Factorise:$1029 - 3x^3$
Answer$1029 - 3x^3$
$= 3(343 - x^3)$
$= 3[(7)^3 - x^3]$
$= 3[(7 - x)(7^2 + 7x + x^2)]$
$= 3(7 - x)(49 + 7x + x^2)$
View full question & answer→Question 812 Marks
Expand:
$(-3a + 4b - 5c)^2$
Answer$(-3a + 4b - 5c)^2 = [(-3a) + (4b) + (-5c)]^2$
$= (-3a)^2 + (4b)^2 + (-5c)^2 + 2(-3a)(4b) + 2(4b)(-5c) + 2(-3a)(-5c)$
$=9a^2 + 16b^2 + 25c^2 - 24ab - 40bc + 30ac$
View full question & answer→Question 822 Marks
Factorise:$a^{12} - b^{12}$
Answer$a^{12} - b^{12}$
$= (a^6)^2 - (b^6)^2$
$= (a^6 - b^6)(a^6 + b^6)$
$= [(a^3)^2 - (b^3)^2][(a^2)^3 + (b^2)^3]$
$= (a^3 - b^3)(a^3 + b^3)[(a^2 + b^2)(a^4 - a^2b^2 + b^4)]$
$= (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2)(a^2 + b^2)(a^4 - a^2b^2 + b^4)$
$= (a - b)(a + b)(a^2 + b^2)(a^2 + ab + b^2)(a^2 - ab + b^2)(a^4 - a^2b^2 + b^4)$
View full question & answer→Question 832 Marks
Factorise:
$\frac{2}{3}\text{x}^2-\frac{17}{3}\text{x}-28$
Answer$\frac{2}{3}\text{x}^2-\frac{17}{3}\text{x}-28$
$=\frac{2}{3}\text{x}^2+\frac{7}{3}\text{x}-8\text{x}-28$
$=\frac{\text{x}}{2}(2\text{x}+7)-4(2\text{x}+7)$
$=\Big(\frac{\text{x}}{3}-4\Big)(2\text{x}+7)$
View full question & answer→Question 842 Marks
Evaluate:$(99)^3$
Answer$(99)^3$
$= (100 - 1)^3$
$= (100)^3 - (1)^3 - 3(100)^2 \times (1) + 3(100)(1)^2$
$= 1000000 - 1 - 30000 + 300$
$= 1000300 - 30001$
$= 970299$
View full question & answer→Question 852 Marks
Factorise:$24x^2 - 41x + 12$
Answer$24x^2 - 41x + 12$
$= 24x^2 - 32x - 9x + 12$
$= 8x(3x - 4) - 3(3x - 4)$
$= (3x - 4)(8x - 3)$
View full question & answer→Question 862 Marks
Factorise:$27x^3 - y^3 - z^3 - 9xyz$
Answer$27x^3 - y^3 - z^3 - 9xyz$
$= (3x)^3 - y^3 - z^3 - 3 \times (3x) \times (-y) \times (-z)$
We know,
$a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$a = 3x, b = -y, c = -z$
$(3x)^3 - y^3 - z^3 - 3 \times (3x) \times (-y) \times (-z)$
$= (3x - y - z)(9x^2 + y^2 + z^2 + 3xy - yz + 3xz)$
View full question & answer→Question 872 Marks
Factorise:$9 - a^2 + 2ab - b^2$
Answer$9 - a^2 + 2ab - b^2$
$= 9 - (a^2 - 2ab + b^2)$
$= 3^2 - (a - b)^2 \big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
$= (3 + a - b)(3 - a + b)$
View full question & answer→Question 882 Marks
Evaluate:$(103)^3$
Answer$(103)^3$
$= (100 + 3)^3$
$= (100)^3 + (3)^3 + 3(100)^2 \times (3) + 3(100)(3)^2$
$= 1000000 + 27 + 90000 + 2700$
$= 1092727$
View full question & answer→Question 892 Marks
Factorise:$x^9 - y^9$
Answer$x^9 - y^9$
$= (x^3)^3 - (y^3)^3$
$= [(x^3 - y^3)][(x^3)^2 + x^3y^3 + (y^3)^2]$
$= [(x - y)(x^2 + xy + y^2)(x^6 + x^3y^3 + y^6)$
View full question & answer→Question 902 Marks
Factorise:$(a - b)^3 + (b - c)^3 + (c - a)^3$
Answer$(a - b)^3 + (b - c)^3 + (c - a)^3$
Putting $(a - b) = x, (b - c) = y$ and $(c - a) = z$
We get: $(a - b)^3 + (b - c)^3 + (c - a)^3$
$= x^3 + y^3 + z^3 [(x + y + z) = (a - b) + (b - c) + (c - a) = 0]$
$= 3xyz [(x + y + z) = 0 $
$\Rightarrow x^3 + y^3 + z^3 = 3xyz]$
$= 3(a - b)(b - c)(c - a)$
View full question & answer→Question 912 Marks
Factorise:$(ax + by)^2 + (bx - ay)^2$
Answer$(ax + by)^2 + (bx - ay)^2$
$= a^2x^2 + b^2y^2 + 2abxy + b^2x^2 + a^2y^2 - 2abxy$
$= a^2x^2 + b^2y^2 + b^2x^2 + a^2y^2$
$= a^2x^2 + b^2x^2 + b^2y^2 + a^2y^2$
$= x^2(a^2 + b^2) + y^2(a^2 + b^2)$
$= (a^2 + b^2)(x^2 + y^2)$
View full question & answer→Question 922 Marks
Factorise:
$\text{x}^2+3\sqrt{3}\text{x}+6$
Answer$\text{x}^2+3\sqrt{3}\text{x}+6$
$=\text{x}^2+2\sqrt{3}\text{x}+\sqrt{3}\text{x}+6$
$=\text{x}(\text{x}+2\sqrt{3})+\sqrt{3}(\text{x}+2\sqrt{3})$
$=(\text{x}+2\sqrt{3})(\text{x}+\sqrt{3})$
View full question & answer→Question 932 Marks
Factorise:$a^3 - 0.064$
Answer$a^3 - 0.064$
$= (a)^3 - (0.4)^3$
$= (a - 0.4)[(a)^2 + a \times 0.4 + (0.4)^2]$ Since $a^3 - b^3 = (a - b)(a^2 + a \times b + b^2)$
$= (a - 0.4)(a^2 + 0.4a + 0.16)$
View full question & answer→Question 942 Marks
Factorise:$x^2 + 11x + 30$
Answer$x^2 + 11x + 30$
$= x^2 + 6x + 5x + 30$
$= x(x + 6) + 5(x + 6)$
$= (x + 6)(x + 5)$
View full question & answer→Question 952 Marks
Factorise:$81x^4 - y^4$
Answer$81x^4 - y^4$
$= (9x^2)^2 - (y^2)^2$
$= (9x^2 - y^2)(9x^2 + y^2)$
$= [(3x)^2 - y^2](9x^2 + y^2)$
$= (3x - y)(3x + y)(9x^2 + y^2)$
View full question & answer→Question 962 Marks
Factorise:$8a^3 + 125b^3 - 64c^3 + 120abc$
Answer$8a^3 + 125b^3 - 64c^3 + 120abc$
$= (2a)^3 + (5b)^3 + (-4c)^3 - 3 \times (2a) \times (5b) \times (-4c)$
$= (2a + 5b - 4c)[(2a)^2 + (5b)^2 + (-4c)^2 - (2a)(5b) - (5b)(-4c) - (2a) \times (-4c)]$
$= (2a + 5b - 4c)(4a^2 + 25b^2 + 16c^2 - 10ab + 20bc + 8ac)$
View full question & answer→Question 972 Marks
Factorise:$(x + 2)^3 + (x - 2)^3$
Answer$(x + 2)^3 + (x - 2)^3$
$= [(x + 2) + (x - 2)][(x + 2)^2 - (x + 2)(x - 2) + (x - 2)^2]$
$= (2x)(x^2 + 4x + 4 - x^2 + 4 + x^2 - 4x + 4)$
$= 2x(x^2 + 12)$
View full question & answer→Question 982 Marks
Factorise:$8a^3 - b^3 - 4ax + 2bx$
Answer$8a^3 - b^3 - 4ax + 2bx$
$= 8a^3 - b^3 - 2x(2a - b)$
$= (2a)^3 - (b)^3 - 2x(2a - b)$ Since $a^3 - b^3 = (a - b)(a^2 + a \times b + b^2)$
$= (2a - b)[(2a)^2 + 2a \times b + (b)^2] - 2x(2a - b)$
$= (2a - b)(4a^2 + 2ab + b^2) - 2x(2a - b)$
$= (2a - b)(4a^2 + 2ab + b^2 - 2x)$
View full question & answer→Question 992 Marks
Factorise:$a^2x^2 + (ax^2 + 1)x + a$
Answer$a^2x^2 + (ax^2 + 1)x + a$
$= a^2x^2 + ax^3 + x + a$
$= ax^2(a + x) + 1(x + a)$
$= (ax^2 + 1)(a + x)$
View full question & answer→Question 1002 Marks
Factorise:$4x^2 - 9y^2 - 2x - 3y$
Answer$4x^2 - 9y^2 - 2x - 3y$
$= (2x)^2 - (3y)^2 - (2x + 3y)$ $\big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
$= (2x + 3y)(2x - 3y) - (2x + 3y)$
$= (2x + 3y)(2x - 3y - 1)$
View full question & answer→