Answer:
[tex]\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }[/tex]
Step-by-step explanation:
Hello,
a and b are the zeros, we can say that
[tex]f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)[/tex]
So we can say that
[tex]a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}[/tex]
Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b
for instance we can write
[tex](x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab[/tex]
and we can notice that
[tex]a^2+b^2=(a+b)^2-2ab[/tex] so
[tex](x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab[/tex]
it comes
[tex]x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}[/tex]
multiply by 3
[tex]3x^2-20x+2*16+5=3x^2-20x+37[/tex]
