Solving Quadratic Equations by Factorising

If two things multiply to give 0, at least one of them must be 0, so set each bracket equal to zero.

$x+3=0$ gives $x=-3$, and $x+4=0$ gives $x=-4$.

The solutions are $x=-3$ and $x=-4$. Notice the signs flip: a $+3$ inside the bracket gives a solution of $-3$.

If two things multiply to give 0, at least one of them must be 0, so set each bracket equal to zero.

$x-2=0$ gives $x=2$, and $x+9=0$ gives $x=-9$.

The solutions are $x=2$ and $x=-9$. Each solution is the opposite sign to the number in its bracket.

For $x^2+bx+c$ we need two numbers that multiply to give $c=10$ and add to give $b=7$. Both conditions must hold.

The pairs multiplying to 10 are 1 and 10 (adding to 11) and 2 and 5 (adding to 7). Only 2 and 5 fit both.

So the factorised form is $(x+2)(x+5)=0$. Checking: $(x+2)(x+5)=x^2+5x+2x+10=x^2+7x+10$.

For $x^2+bx+c$ we need two numbers that multiply to give $c=28$ and add to give $b=-11$. Since the product is positive and the sum is negative, both numbers must be negative.

The pairs multiplying to 28 are $-1$ and $-28$, $-2$ and $-14$, and $-4$ and $-7$. Only $-4$ and $-7$ add to $-11$.

So the factorised form is $(x-4)(x-7)=0$. Checking: $(x-4)(x-7)=x^2-7x-4x+28=x^2-11x+28$.

For $x^2+bx+c$ we need two numbers that multiply to give $c=-18$ and add to give $b=-3$. Since the product is negative, one number is positive and one is negative.

The pairs multiplying to $-18$ include $-9$ and 2 (adding to $-7$), $-6$ and 3 (adding to $-3$) and $-3$ and 6 (adding to 3). Only $-6$ and 3 fit both conditions.

So the factorised form is $(x-6)(x+3)=0$. Checking: $(x-6)(x+3)=x^2+3x-6x-18=x^2-3x-18$.

Every term has a factor of 5, so divide the whole equation by 5: $x^2+6x-16=0$.

Now find two numbers that multiply to $-16$ and add to 6. These are $-2$ and 8, so $(x-2)(x+8)=0$.

Setting each bracket to zero gives $x-2=0$, so $x=2$, and $x+8=0$, so $x=-8$. The solutions are $x=2$ and $x=-8$.