Answer:
[tex]\large\boxed{(x^2-3x)(4x^2+2x-9)=4x^4-10x^3-15x^2+27x}[/tex]
Step-by-step explanation:
[tex](x^2-3x)(4x^2+2x-9)\qquad\text{use}\ (a+b)(c+d+e)=ac+ad+ae+bc+bd+be\\\\=(x^2)(4x^2)+(x^2)(2x)+(x^2)(-9)+(-3x)(4x^2)+(-3x)(2x)+(-3x)(-9)\\\\=4x^4+2x^3-9x^2-12x^3-6x^2+27x\qquad\text{combine like terms}\\\\=4x^4+(2x^3-12x^3)+(-9x^2-6x^2)+27x\\\\=4x^4-10x^3-15x^2+27x[/tex]