Question Multiply: (6x−5y)(−3x+2y)(6x−5y)(−3x+2y) (1 point) Responses −18x2+27xy−10y2−18x2+27xy−10y2negative 18 x squared plus 27 x y minus 10 y squared 3x−3y3x−3y3 x minus 3 y −18x2+3xy+10y2−18x2+3xy+10y2negative 18 x squared plus 3 x y plus 10 y squared −31x2y2

Answer 1

To find the product $(6x-5y)(-3x+2y)(6x-5y)(-3x+2y)$, we first multiply the two binomials and then square the result.

1. Multiply $(6x-5y)$ and $(-3x+2y)$: $$(6x-5y)(-3x+2y)=6x(-3x)+6x(2y)-5y(-3x)-5y(2y)$$ $$=-18x^2+12xy+15xy-10y^2$$ $$=-18x^2+27xy-10y^2$$

2. Now square the result: $$(-18x^2+27xy-10y^2{)}^2$$

Using the square of a trinomial formula $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2ac+2bc$, where $a=-18x^2$, $b=27xy$, and $c=-10y^2$:

• $a^2=(-18x^2{)}^2=324x^4$
• $b^2=(27xy{)}^2=729x^2{y}^2$
• $c^2=(-10y^2{)}^2=100y^4$
• $2ab=2(-18x^2)(27xy)=-972x^{3}y$
• $2ac=2(-18x^2)(-10y^2)=360x^2{y}^2$
• $2bc=2(27xy)(-10y^2)=-540x{y}^3$

Now combine these results: $$(-18x^2+27xy-10y^2{)}^2=324x^4-972x^{3}y+(729+360)x^2{y}^2-540x{y}^3+100y^4$$ $$=324x^4-972x^{3}y+1089x^2{y}^2-540x{y}^3+100y^4$$

Thus, the final answer is: $$324x^4-972x^{3}y+1089x^2{y}^2-540x{y}^3+100y^4$$

However, the choices you provided do not include this complete polynomial. Based on your initial multiplication, the answer from the options must be recognized.

None of the provided choices directly match the full expansion result from above, but checking back, the first multiplication yielded the principal product of $-18x^2+27xy-10y^2$, which confirms correct accounting in our calculation.

Therefore, based on the step 1 results, the correct simplified result is: $$\boxed{{\displaystyle -18x^2+27xy-10y^2}}$$