Question

multiply the following algebraic expression:-
12х + 9;2x^2-3x+ 6​

Answer 1

Simplify (4x2 – 4x – 7)(x + 3)

Here's what the multiplication looks like when it's done horizontally:

(4x2 – 4x – 7)(x + 3)

(4x2 – 4x – 7)(x) + (4x2 – 4x – 7)(3)

4x2(x) – 4x(x) – 7(x) + 4x2(3) – 4x(3) – 7(3)

4x3 – 4x2 – 7x + 12x2 – 12x – 21

4x3 – 4x2 + 12x2 – 7x – 12x – 21

4x3 + 8x2 – 19x – 21

That was painful! Now I'll do it vertically:

4x^2 – 4x – 7 is positioned above x + 3; first row: +3 times –7 is –21, carried down below the +3; +3 times –4x is –12x, carried down below the x; +3 times 4x^2 is +12x^2, carried down to the left of the –12x; second row: x times –7 is –7x, carried down below the –12x; x times –4x is –4x^2, carried down below the +12x^2; x times 4x^2 is 4x^3, carried down to the left of the –4x^2; adding down: 4x^3 + (+12x^2) + (–4x^2) + (–12x) + (–7x) + (–21) = 4x^3 + 8x^2 – 19x – 21

That was a lot easier! But, by either method, the answer is the same:

4x3 + 8x2 – 19x – 21

Answer 2

Answer:

4x^3+8x^2-19x-21

good night