How do you solve #4y^3-2=y-8y^2# by factoring and then using the zero-product principle?
3 Answers
Explanation:
The given eqn. is,
These roots satisfy the given eqn.
Hence, the
Explanation:
#"rearrange and equate to zero"#
#rArr4y^3+8y^2-y-2=0larrcolor(blue)"factorise by grouping"#
#color(red)(4y^2)(y+2)color(red)(-1)(y+2)=0#
#rArr(y+2)(color(red)(4y^2-1))=0#
#4y^2-1color(blue)" is a difference of squares"#
#•color(white)(x)a^2-b^2=(a-b)(a+b)#
#rArr(y+2)(2y-1)(2y+1)=0#
#"equate each factor to zero (zero-product principle) and"#
#"solve for y"#
#y+2=0rArry=-2#
#2y-1=0rArry=1/2#
#2y+1=0rArry=-1/2#
Explanation:
First, we put all the terms on one side.
Then, we group the factorize it by using common factor.
Recall that the zero-product principle tells, when
Reminder:
Here is the answer :)
Hope it can help you.