how do i factorise these two questions?

1)
4a^2c + 8ac^3
= 4(a^2c + 2ac^3)
= 4ac(a + 2c^2)

2)
y^2 - 11y - 24
= y^2 - 11y/2 - 11y/2 + 121/4 - 24 - 121/4
= (y^2 - 11y/2) - (11y/2 - 121/4) - 96/4 - 121/4
= y(y - 11/2) - 11/2(y - 11/2) - 217/4
= (y - 11/2)(y - 11/2) - 217/4
= (y - 11/2)^2 - 217/4

2.
Use completing the square to factorise this expression because it contains irrational roots.
y^2-11y -24
= y^2-11y+(-11/2)^2-(-11/2)^2-24
= [ y^2-11y+(-11/2)^2]-(-11/2)^2-24
= [ y^2-11y+(-11/2)^2]-(121/4)-24
= [ y^2-11y+(-11/2)^2]-(121/4)-(96/4)
= [ y^2-11y+(-11/2)^2]-(217/4)
= (y-11/2)^2-[(√217)/2]^2
= {y-11/2-[(√217)/2]}{y+11/2+[(√217)/2]}
= {y-11/2-[(√217)/2]}{y+11/2+[(√217)/2]}
= {y-[(11+√217)/2]}{y+[(11+√217)/2]}

1. If you factor 4a^2c+8ac^3 completely, the answer is
4ac(a+2c^2).So, your answer is correct.

2. Is your expression for # 2 correct? I think it's y^2-10y-24. It's a general trinomial. So, the factors are (y-12) (y+2)..

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Question 1
4ac (a + 2c²) is correct

Question 2
Could it be that you have a wrong sign ?
Could it be :-
y² - 11y + 24
(y - 8)(y - 3)

OR

y² - 11y - 24
(y² - 11y + 121/4) - 24 - 121/4
(y - 11/2)² - 217/4

1. Your answer is fine - your calculator isn't up to it !
2. Are you sure you've got the signs right ?

4a^2c+8ac^3
=4ac(a+2c^2)


y^2-11y-24
= y^2(-8y+3y) -24
=(y^2-8y)+(3y-24)
=y(y-8)+3(y-8)
=(y+3)(y-8)

1. your way will not work out because your your ans, 4ac is not supposed to have an a or c. It cannot have an a because the a in 4a^2c has a power attached to it so you cannot grp that with other common terms while there is no c in 4a^2c. c is part of the power not part of the main number.

2. Both signs cannot be minus, so Im not sure

1 your way is actually correct and i dont know whats with your calculator.

2. did your calculator work it out in decimals? or no answer?

if you used the quadratic equation it would become

(11+root 217)/2 or (11-root217)/2

or 12.8654 or -1.8654