Question

(2x² + 3x-7) x(3x²- 5x+4)
please give full explanation​

Answer 1

Simplify (3x2 – 9x + 5)(2x2 + 4x – 7)

I'll take my time, and do my work neatly:

3x^2 – 9x + 5 is over 2x^2 + 4x – 7; first row: –7 times +5 is – 35, to below –7; –7 times –9x is +63x, to below +4x; –7 times 3x^2 is –21x^2, to below 2x^2; second row: +4x times +5 is +20x, to below +63x; +4x times –9x is –36x^2, to below –21x^2; +4x times 3x^2 is +12x^3, to left of –36x^2; third row: 2x^2 times +5 is +10x^2, to below –36x^2; 2x^2 times –9x is –28x^3, to below +12x^3; 2x^2 times 3x^2 is 6x^4, to left of –18x^3; adding down: 6x^4 + 12x^3 – 18x^3 – 21x^2 – 36x^2 + 10x^2 + 63x + 20x – 35 = 6x^4 – 6x^3 – 47x^2 + 83x – 35

6x4 – 6x3 – 47x2 + 83x – 35

Answer 2

Answer:

By using horizontal method,

We have;

(2x2+ 3x – 7) ×(3x2 – 5x + 4)

= 2x2(3x2 – 5x + 4) + 3x(3x2 – 5x + 4) – 7(3x2 – 5x + 4)

= 6x4 – 10x3 + 8x2 + 9x3 – 15x2 + 12x– 21x2 + 35x – 28

Now, putting equal power terms together,

we get,

= 6x4 – 10x3 + 9x3 + 8x2 – 15x2– 21x2 + 35x + 12x– 28

= 6x4 – x3 – 28x2 + 47x – 28