3) for any factorable trinomial x^2+bx+c, will the absolute value of b sometimes,always, or never be less than the absolute value of c? explain

idk

4) give an example of a trinomial that can be factored using the factoring techniques presented. then factor the trinomial. idk

5) factor (4y-5)^2 + 3(4y-5)-70

i got this

Make the simplifying substitution
z = (4y-5):
z^2 + 3z - 70
Factor:
(z + 10) (z - 7)
Undo the substitution:
((4y-5) + 10) ((4y-5) - 7)
Combine like terms:
(4y + 5) (4y - 12)

can you help me this

3) sometimes

4) what techniques were presented?
5) correct

3) why is it sometmes

4) idk

(x - 2)(x + 8) b is less than c

(x - 5)(2x + 1) b is greater than c

3) To determine if the absolute value of b is sometimes, always, or never less than the absolute value of c for any factorable trinomial x^2+bx+c, we need to consider the discriminant.

The discriminant is the term inside the square root in the quadratic formula, given by b^2 - 4ac. In this case, a=1, b=b, and c=c.

If the discriminant is positive (b^2 - 4ac > 0), then the quadratic equation has two distinct real roots. In this situation, the absolute value of b can be either greater than or less than the absolute value of c. Therefore, the absolute value of b is sometimes less than the absolute value of c.

If the discriminant is zero (b^2 - 4ac = 0), then the quadratic equation has one real root with multiplicity 2. In this case, the absolute value of b is always equal to the absolute value of c.

If the discriminant is negative (b^2 - 4ac < 0), then the quadratic equation has two complex roots. In this situation, the absolute value of b is always greater than the absolute value of c.

Therefore, the absolute value of b can be sometimes less than, always equal to, or never less than the absolute value of c, depending on the nature of the roots of the quadratic equation.

4) To give an example of a trinomial that can be factored using the factoring techniques presented, let's consider the trinomial x^2 + 5x + 6.

To factor this trinomial, we need to find two numbers that multiply to give 6 and add up to 5. In this case, the numbers are 2 and 3.

Therefore, the factored form of x^2 + 5x + 6 is (x + 2)(x + 3).

5) To factor the expression (4y-5)^2 + 3(4y-5) - 70, you correctly made the substitution z = (4y-5) to simplify the expression.

The expression becomes z^2 + 3z - 70.

Now, we can factor this simplified expression. The factors of -70 that add up to 3 are 10 and -7. Therefore, we can write the expression as:

(z + 10)(z - 7).

Since we made the substitution z = (4y-5), we can now undo the substitution:

((4y-5) + 10)((4y-5) - 7).

Finally, we can combine like terms:

(4y + 5)(4y - 12).

So, the factored form of (4y-5)^2 + 3(4y-5) - 70 is (4y + 5)(4y - 12).

Please go back and see my last comment on your first questions. I got distracted and gave you some incorrect info. Sorry.