Here are the detailed step-by-step solutions to the questions.
We need two numbers that multiply to give $+12$ and add to give $+7$. These numbers are $3$ and $4$.
$(x+3)(x+4) = 0$
For the product to be zero, one of the brackets must be zero.
$x+3=0$ or $x+4=0$
Answer: $x = -3$ or $x = -4$
We need two numbers that multiply to give $-10$ and add to give $+3$. These numbers are $5$ and $-2$.
$(x+5)(x-2) = 0$
$x+5=0$ or $x-2=0$
Answer: $x = -5$ or $x = 2$
We need two numbers that multiply to give $+15$ and add to give $-8$. These numbers are $-3$ and $-5$.
$(x-3)(x-5) = 0$
$x-3=0$ or $x-5=0$
Answer: $x = 3$ or $x = 5$
We need two numbers that multiply to give $-24$ and add to give $-5$. These numbers are $-8$ and $3$.
$(x-8)(x+3) = 0$
$x-8=0$ or $x+3=0$
Answer: $x = 8$ or $x = -3$
This is a special case called the 'difference of two squares', $a^2 - b^2 = (a-b)(a+b)$. Here, $a=x$ and $b=7$.
$(x-7)(x+7) = 0$
$x-7=0$ or $x+7=0$
Answer: $x = 7$ or $x = -7$
Since the coefficient of $x^2$ is not 1, we look for two numbers that multiply to give $a \times c$ (which is $2 \times 3 = 6$) and add to give $b$ (which is $7$). The numbers are $6$ and $1$. We split the middle term.
$2x^2 + 6x + 1x + 3 = 0$
Factor by grouping:
$2x(x+3) + 1(x+3) = 0$
$(2x+1)(x+3) = 0$
$2x+1=0$ or $x+3=0$
Answer: $x = -1/2$ or $x = -3$
We need two numbers that multiply to give $a \times c$ ($3 \times 4 = 12$) and add to give $b$ ($-13$). The numbers are $-12$ and $-1$. We split the middle term.
$3x^2 - 12x - 1x + 4 = 0$
Factor by grouping:
$3x(x-4) - 1(x-4) = 0$
$(3x-1)(x-4) = 0$
$3x-1=0$ or $x-4=0$
Answer: $x = 1/3$ or $x = 4$
First, rearrange the equation so it equals zero: $x^2 + 10x + 21 = 0$. Now we need two numbers that multiply to give $21$ and add to give $10$. These are $7$ and $3$.
$(x+7)(x+3) = 0$
$x+7=0$ or $x+3=0$
Answer: $x = -7$ or $x = -3$
We need two numbers that multiply to give $a \times c$ ($4 \times -5 = -20$) and add to give $b$ ($8$). The numbers are $10$ and $-2$. We split the middle term.
$4x^2 + 10x - 2x - 5 = 0$
Factor by grouping:
$2x(2x+5) - 1(2x+5) = 0$
$(2x-1)(2x+5) = 0$
$2x-1=0$ or $2x+5=0$
Answer: $x = 1/2$ or $x = -5/2$
We need two numbers that multiply to give $a \times c$ ($6 \times -2 = -12$) and add to give $b$ ($-1$). The numbers are $-4$ and $3$. We split the middle term.
$6x^2 - 4x + 3x - 2 = 0$
Factor by grouping:
$2x(3x-2) + 1(3x-2) = 0$
$(2x+1)(3x-2) = 0$
$2x+1=0$ or $3x-2=0$
Answer: $x = -1/2$ or $x = 2/3$