tertre

We are asked to find: $x^2+y^2$, for $x$ and $y$ are the roots of the equation.

$3x^2+4x+12$ has a sum of $\frac{-4}{3}.$

$3x^2+4x+12$ has a product of $\frac{12}{3}=4.$

Use the manipulation:$(x+y)^2-2(xy)=x^2+y^2$.

Thus, the answer is $(\frac{-4}{3})^2-2(4)=\frac{16}{9}-\frac{72}{9}=\frac{-56}{9}.$

1. Not sure if this is the fastest way.

(l+2)(h+2)(w+2)=lwh+2(lh+lw+wh)+4(l+w+h)+8

120+2(80)+4(15)+8=348

2. Hint: The base is an equilateral triangle. Try to drop the altitude.

This involves factoring from Vieta's Formulas:

The first step is to break $a^3+b^3+c^3$ into $3r_1r_2r_3+(r_1+r_2+r_3)[r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1)$.

The confusing part might be trying to find $r_1^2+r_2^2+r_3^2$, yet we know that is equal to $(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).$

This can be better written as $3r_1r_2r_3+(r_1+r_2+r_3)[(r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1)].$

Remember that $r_1, r_2$, and $r_3$ are the roots of the polynomial which is an expression.

Note that $r_1+r_2+r_3=\frac{-b}{a}$ , $r_1r_2+r_2r_3+r_3r_1=\frac{c}{a}$, and  $r_1r_2r_3=\frac{-d}{a}.$

Try to plug the values in, and be careful and don't forget the $x^2$ term.

This is a typical problem for the Rational Root Theorem: Read more here: https://brilliant.org/wiki/rational-root-theorem/

Since the prime factorization of 42=2*3*7, but we can also add the value one(1).

There could be negative values, too. I agree with Melody.

A faster way is the Euler's Theorem.

phi26=12 and 15^145phi12=15^1=15.

Testing the values with y=1, y=2, y=3, y=4 yields negative value by using Synthetic Division. Yet, we can realize something. The -25x^3 is really messing up the positive values, so we need to find a value larger than 25. Since 6*4=24, that won't work, yet y=7 will work since 7*4=28, which is greater than 25.And yes, y=7 does work.

I'm not sure if this is correct, but my angle relationships,

5x-7=2x+11

3x=18

x=6

Thus, angle ABP and angle EDG both meaure twenty-three(23) degrees.

The angle ABD measures 180-23=157degrees.

Posted two responses at once.

We try the values y= 1, 2, 3, 4, 5, 6,...and so on.

For the function to have an upper bound, the coefficients of the polynomial when the original polynomial divided by (y) has to be positive. I recommend using Sythetic Division to divide the polynomial with the number.

Trying y=1, gives us a polynomial that alternates in signs. Same with y=2. Yet, y=3 gives us a polynomial with all positive coefficients. y=3: Answer

Yeah, post the image properly, please. Just posting the asymptote in the question box won't show up.