Question

x2-3x-28÷(x2+10x+24)(x-5)

Answer 1

x²-3x-28÷(x²+10x+24)(x-5)​

$\frac{x^2-3x-28}{(x^2+10x+24)(x-5)}$

First, let's factorise x²-3x-28

x²-3x-28

x²-7x+4x-28

x(x-7) + 4(x-7)

(x-7)(x+4)

Next

We factorise; x² + 10x +24

x²+ 10x + 24

x² + 6x + 4x + 24

x(x+6) + 4(x+6)

(x+6)(x+4)

Now; let's plug them back into our original expression

$\frac{(x-7)(x+4)}{(x+6)(x+4)(x-5)}$

(x+4) at the numerator will cancel-out (x+4) at the denominator

Hence;

$\frac{x-7}{(x+6)(x-5)}$

we can leave the answer like this or open the bracket and then simplify

That is;

(x+6)(x-5) = x² -5x +6x - 30 = x² +x - 30

Plugging it back yields;

$\frac{x-7}{x^2+x\text{ -30}}$