Answer:
![(x-3)(x-9)(x-4)=x^3-16x^2+75x-108](https://img.qammunity.org/2021/formulas/mathematics/high-school/sn08emyzbh2d7ciwdmpvvbzuddmezzmms4.png)
Explanation:
Given:
The expression to expand is given as:
![(x-3)(x-9)(x-4)](https://img.qammunity.org/2021/formulas/mathematics/high-school/4lg0oas7nthpezhx3gpf47haxjx68zd8vf.png)
Let us expand the first two binomials of the given expression using FOIL method.
The FOIL method states that:
![(a + b)(c + d) = ac + ad + bc + bd](https://img.qammunity.org/2021/formulas/mathematics/high-school/glh2gy27z24k9alx4j3qbb9q62jd36vm10.png)
![(x-3)(x-9)=(x* x)+(x* -9)+(-3* x)+(-3* -9)\\\\(x-3)(x-9)=x^2-9x-3x+27=x^2-12x+27](https://img.qammunity.org/2021/formulas/mathematics/high-school/63omc650lerf2166ylnft8c8lm9mtab6j9.png)
Now, let us multiply the result with the remaining binomial. Multiplying each term of the trinomial with each term of the binomial, we get:
![(x^2-12x+27)(x-4)\\\\= (x^2* x)+ (x^2* -4)+(-12x* x)+(-12x* -4)+(27* x)+(27* -4)\\\\=x^3-4x^2-12x^2+48x+27x-108\\\\=x^3-16x^2+75x-108...........(\because-12x^2-4x^2=-16x^2,48x+27x=75x)](https://img.qammunity.org/2021/formulas/mathematics/high-school/7uh4e2cs4fhsoztyplp4etv33pmx8cfi8b.png)
Therefore, the equivalent expression after expanding is given as:
![(x-3)(x-9)(x-4)=x^3-16x^2+75x-108](https://img.qammunity.org/2021/formulas/mathematics/high-school/sn08emyzbh2d7ciwdmpvvbzuddmezzmms4.png)