Expanding brackets and simplifying

(a) $3 \times 4t = 12t$

(b) $8a - 3a = 5a$, then $5a + 2a = 7a$

(a) $3m - m - m + 3m = (3 - 1 - 1 + 3)m = 4m$

(b) Multiply the numbers: $2 \times 4 = 8$, then write the letters: $8np$.

$7a + a = 8a$

$8a - 5a = 3a$

(a) The terms are all $h^3$: $8 + 14 - 2 = 20$, so $20h^3$.

(b) $9y + 12y = 21y$, and $21y \div 3 = 7y$.

(a) Multiply the numbers and write the letters together: $2 \times 5 = 10$, so $10ab$.

(b) $x$ terms: $3x + 5x = 8x$

$y$ terms: $2y - y = y$

Answer: $8x + y$

(a) $\frac{15a}{3} = 5a$

(b) $b$ terms: $5b - 7b = -2b$. $c$ terms: $4c + c = 5c$.

So $19 - 2b + 5c$

(c) The highest common factor of 8 and 6 is 2: $8d - 6 = 2(4d - 3)$

(a) $3 \times 4 = 12$ and $3 \times -2x = -6x$, giving $12 - 6x$

(b) $\frac{3y}{4} = 12$, so $3y = 48$ and $y = 16$

(c) The highest common factor of 4 and 6 is 2: $4p + 6 = 2(2p + 3)$

(a) $y \times y + y \times 5 = y^2 + 5y$

(b) The highest common factor of 4 and 6 is 2: $2(2a - 3)$

(c) $10x - 8 = 21$, so $10x = 29$ and $x = 2.9$

(d) $4 \times 5 = 20$, $e^2 \times e = e^3$, $f \times f^3 = f^4$, giving $20e^3f^4$

(a) $5 \times 2m = 10m$ and $5 \times (-3) = -15$, so $10m - 15$

(b) 3 is a factor of both terms: $3n + 12 = 3(n + 4)$

(a) Write the number first, then the letters: $7ab$

(b) $y$ multiplied by itself three times is $y^3$

(c) Top: $e \times e \times e \times f = e^3 f$. Bottom: $e \times e \times f \times f = e^2 f^2$

Cancel two $e$s and one $f$: $\frac{e^3 f}{e^2 f^2} = \frac{e}{f}$

(a) $6y - 15 + 7y + 14 = 13y - 1$

(b) The HCF of $6x^2$ and $15x$ is $3x$: $6x^2 + 15x = 3x(2x + 5)$

(c) $f - 11 = 3g$, so $g = \frac{f - 11}{3}$

(a) $6y - 15 + 7y + 14 = 13y - 1$

(b) The highest common factor of $6x^2$ and $15x$ is $3x$, so $6x^2 + 15x = 3x(2x + 5)$.

(c) $f - 11 = 3g$, so $g = \frac{f - 11}{3}$.

(a) Multiply the powers: $(m^2)^3 = m^{2 \times 3} = m^6$

(b) Add the powers: $x^{5+8} = x^{13}$

(c) $4p \times p^2 = 4p^3$ and $4p \times 3p = 12p^2$, so $4p^3 + 12p^2$

(a) Multiply the powers: $(x^3)^5 = x^{15}$

(b) $4x + 12 + 28 - 14x = 40 - 10x$

(c) The HCF of $15x^3$ and $3x^2y$ is $3x^2$.

$15x^3 + 3x^2y = 3x^2(5x + y)$

Subtracting a bracket changes the sign of every term inside it:

$-(x^2 - 2x - 4) = -x^2 + 2x + 4$

So the answer should be $x^2 + 3x + 5 - x^2 + 2x + 4 = 5x + 9$. Peter kept $-2x - 4$.

(a) $x \times x = x^2$, $x \times (-9) = -9x$, $5 \times x = 5x$, $5 \times (-9) = -45$

$x^2 - 9x + 5x - 45 = x^2 - 4x - 45$

(b) The highest common factor of $9x^2$ and $6x$ is $3x$: $3x(3x + 2)$

$5(p + 3) = 5p + 15$

$-2(1 - 2p) = -2 + 4p$

$5p + 15 - 2 + 4p = 9p + 13$

Multiply each term in the first bracket by each term in the second:

$x \times x = x^2$, $x \times (-1) = -x$, $3 \times x = 3x$, $3 \times (-1) = -3$

$x^2 - x + 3x - 3 = x^2 + 2x - 3$

(a) $(3x + 2)(2x - 5) = 6x^2 - 15x + 4x - 10 = 6x^2 - 11x - 10$

(b) $x^2 - 16 = x^2 - 4^2 = (x - 4)(x + 4)$

(a) Anything (non-zero) to the power 0 is 1.

(b) Cancel one factor of $(x - 4)$ from top and bottom: $\frac{8(x-4)}{(x-4)^2} = \frac{8}{x-4}$.

(c) Cube each part: $3^3 = 27$, $(n^4)^3 = n^{12}$, $(w^2)^3 = w^6$, giving $27n^{12}w^6$.

(a) Multiply the powers: $(p^2)^5 = p^{2 \times 5} = p^{10}$

(b) Divide the numbers and subtract the powers:

$12 \div 6 = 2$, $x^{7-3} = x^4$, $y^{3-1} = y^2$

Answer: $2x^4y^2$

(a) $(5x + 2)(2x - 3) = 10x^2 - 15x + 4x - 6 = 10x^2 - 11x - 6$

(b) Two numbers that multiply to 3 and add to 4: 1 and 3.

$x^2 + 4x + 3 = (x + 1)(x + 3)$

$(3x - 1)(2x + 3) = 6x^2 + 9x - 2x - 3 = 6x^2 + 7x - 3$

$(6x^2 + 7x - 3)(x - 5) = 6x^3 - 30x^2 + 7x^2 - 35x - 3x + 15$

$= 6x^3 - 23x^2 - 38x + 15$

Numerator: $(6x^5y^3)^2 = 36x^{10}y^6$

Denominator: $3x^2y^7 \times 4xy^{-3} = 12x^3y^4$

Divide: $\frac{36}{12} = 3$, $x^{10-3} = x^7$, $y^{6-4} = y^2$

Answer: $3x^7y^2$

$(x + 7)(x - 2) = x^2 + 5x - 14$

$(x^2 + 5x - 14)(x + 3) = x^3 + 3x^2 + 5x^2 + 15x - 14x - 42$

$= x^3 + 8x^2 + x - 42$

$(x-1)(x+3) = x^2 + 3x - x - 3 = x^2 + 2x - 3$

$(x^2 + 2x - 3)(x - 5) = x^3 - 5x^2 + 2x^2 - 10x - 3x + 15$

$= x^3 - 3x^2 - 13x + 15$, so $a = 1$, $b = -3$, $c = -13$, $d = 15$, all integers.

(a) $3^4 = 81$, $(x^5)^4 = x^{20}$, $(y^6)^4 = y^{24}$, so $81x^{20}y^{24}$

(b) $(x+2)(x-3) = x^2 - x - 6$

$(x^2 - x - 6)(x + 4) = x^3 + 4x^2 - x^2 - 4x - 6x - 24 = x^3 + 3x^2 - 10x - 24$

Split into a top rectangle $(x + 1)$ by $4$ and a bottom rectangle $(2x + 6)$ by $(x + 11) - 4 = x + 7$.

Top: $4(x + 1) = 4x + 4$

Bottom: $(2x + 6)(x + 7) = 2x^2 + 14x + 6x + 42 = 2x^2 + 20x + 42$

$A = 2x^2 + 24x + 46$

$(x - 2)(3x + 2) = 3x^2 + 2x - 6x - 4 = 3x^2 - 4x - 4$

$(3x^2 - 4x - 4)(2x + 3) = 6x^3 + 9x^2 - 8x^2 - 12x - 8x - 12$

$= 6x^3 + x^2 - 20x - 12$

$(3x + 2)(2x + 1) = 6x^2 + 7x + 2$

$(6x^2 + 7x + 2)(x - 5) = 6x^3 - 30x^2 + 7x^2 - 35x + 2x - 10$

$= 6x^3 - 23x^2 - 33x - 10$

$(3x - 1)(x + 5) = 3x^2 + 15x - x - 5 = 3x^2 + 14x - 5$

$(3x^2 + 14x - 5)(4x - 3)$

$= 12x^3 - 9x^2 + 56x^2 - 42x - 20x + 15$

$= 12x^3 + 47x^2 - 62x + 15$

(a) $(2x+1)(x+3) = 2x^2 + 7x + 3$

$(2x^2 + 7x + 3)(3x + 7) = 6x^3 + 14x^2 + 21x^2 + 49x + 9x + 21 = 6x^3 + 35x^2 + 58x + 21$

So $a = 6, b = 35, c = 58, d = 21$, all integers.

(b) $-\frac{3}{5} < 1 - x < \frac{3}{5}$

$\frac{2}{5} < x < \frac{8}{5}$

(a) $(x - 2)(2x + 3) = 2x^2 - x - 6$

$(2x^2 - x - 6)(x + 1) = 2x^3 + 2x^2 - x^2 - x - 6x - 6 = 2x^3 + x^2 - 7x - 6$

(b) $4 + n - 2 = -3$, so $n = -5$

(c) $x = \frac{4 \pm \sqrt{16 + 60}}{10} = \frac{4 \pm \sqrt{76}}{10}$, so $x = 1.27$ or $x = -0.472$