3 years ago

Multiply (3x^2-2x)(2x^2+3x-1)

Mathematics

2 answers:

Vladimir [108]3 years ago

6 0

Answer:

$\large\boxed{(3x^2-2x)(2x^2+3x-1)==6x^4+5x^3-8x^2+2x}$

Step-by-step explanation:

Use the distributive property and $(a^n)(a^m=)=a^{n+m}$

$(3x^2-2x)(2x^2+3x-1)\\\\=(3x^2)(2x^2)+(3x^2)(3x)+(3x^2)(-1)+(-2x)(2x^2)+(-2x)(3x)+(-2x)(-1)\\\\=6x^4+9x^3-3x^2-4x^3-6x^2+2x\qquad\text{combine like terms}\\\\=6x^4+(9x^3-4x^3)+(-3x^2-5x^2)+2x\\\\=6x^4+5x^3-8x^2+2x$

____ [38]3 years ago

3 0

Answer:

6x^4+5x^3-9x^2+2x

Step-by-step explanation:

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Multiply (3x^2-2x)(2x^2+3x-1)​

Multiply (3x^2-2x)(2x^2+3x-1)