ax^3+bx^2+cx = a(x^3+bx^2/a+c/a)
In this case, a=8, b=a, and c=−9a
3
. So, the factored expression is
8x^3+ax^2-9a^3 = 8(x^3+(a/8)x^2+(-9a^3/8))
x^3+(a/8)x^2+(-9a^3/8) = x(x^2+(a/8)x-(9a^3/8))
The quadratic expression x
2
+(a/8)x−(9a
3
/8) can be factored as (x−3a)(x+3a) using the sum-product pattern.
=8(x-3a)(x+3a)
In this case, a=8, b=a, and c=−9a
3
. So, the factored expression is
8x^3+ax^2-9a^3 = 8(x^3+(a/8)x^2+(-9a^3/8))
x^3+(a/8)x^2+(-9a^3/8) = x(x^2+(a/8)x-(9a^3/8))
The quadratic expression x
2
+(a/8)x−(9a
3
/8) can be factored as (x−3a)(x+3a) using the sum-product pattern.
=8(x-3a)(x+3a)
Original post by MrsMia
8x^3+ax^2-9a^3
Can someone pls show me how to factorise this? 😭
Can someone pls show me how to factorise this? 😭
What tools do you have at your disposal? For instance, have you learned remainder/factor theorem?
If so... First step in spoilers:
Spoiler
Usually if you want to factorize a cubic, factor theorem is your friend.
(edited 8 months ago)
Yes I have learnt factor and remainder theorem but how do we use it here?
Original post by tonyiptony
What tools do you have at your disposal? For instance, have you learned remainder/factor theorem?
If so... First step in spoilers:
Usually if you want to factorize a cubic, factor theorem is your friend.
If so... First step in spoilers:
Spoiler
Usually if you want to factorize a cubic, factor theorem is your friend.
Original post by MrsMia
Yes I have learnt factor and remainder theorem but how do we use it here?
Read the hint in spoilers, see if you can make any headway.
Note: the thing you want to find is not complicated. But maybe as a minor help if you are still stuck, "a" here is constant.
Original post by khalumbadavid
ax^3+bx^2+cx = a(x^3+bx^2/a+c/a)
In this case, a=8, b=a, and c=−9a
3
In this case, a=8, b=a, and c=−9a
3
EDIT your post - you are breaking forum rules by posting a solution.
Original post by khalumbadavid
ax^3+bx^2+cx = a(x^3+bx^2/a+c/a)
In this case, a=8, b=a, and c=−9a
3
. So, the factored expression is
8x^3+ax^2-9a^3 = 8(x^3+(a/8)x^2+(-9a^3/8))
x^3+(a/8)x^2+(-9a^3/8) = x(x^2+(a/8)x-(9a^3/8))
The quadratic expression x
2
+(a/8)x−(9a
3
/8) can be factored as (x−3a)(x+3a) using the sum-product pattern.
=8(x-3a)(x+3a)
In this case, a=8, b=a, and c=−9a
3
. So, the factored expression is
8x^3+ax^2-9a^3 = 8(x^3+(a/8)x^2+(-9a^3/8))
x^3+(a/8)x^2+(-9a^3/8) = x(x^2+(a/8)x-(9a^3/8))
The quadratic expression x
2
+(a/8)x−(9a
3
/8) can be factored as (x−3a)(x+3a) using the sum-product pattern.
=8(x-3a)(x+3a)
Spoiler
your first part 'cx' why? there is no coefficient of x in the original.
Original post by MrsMia
Yes I have learnt factor and remainder theorem but how do we use it here?
one simple thing is the observation of the known coefficients you have, 8 1 -9.
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