How do you solve #4y^3-2=y-8y^2# by factoring and then using the zero-product principle?

3 Answers
Dec 11, 2017

#"Soln. Set="{-2,-1/2,1/2}"#.

Explanation:

The given eqn. is, #4y^3-2=y-8y^2.#

#:. 4y^3-2+8y^2-y=0.#

#:. ul(4y^3+8y^2)-ul(y-2)=0.#

#:. 4y^2(y+2)-1(y+2)=0.#

#:. (y+2)(4y^2-1)=0.#

#:. (y+2)(2y+1)(2y-1)=0.#

#:. y+2=0, or 2y+1=0, or 2y-1=0.#

#:. y=-2, or y=-1/2, or y=1/2.#

These roots satisfy the given eqn.

Hence, the #"Soln. Set="{-2,-1/2,1/2}"#.

Dec 11, 2017

#y=-2,y=+-1/2#

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#

Dec 11, 2017

#y=-2 or y=1/2 or y=-1/2#

Explanation:

First, we put all the terms on one side.
#4y^3-2=y-8y^2#
#4y^3+8y^2-y-2=0#

Then, we group the factorize it by using common factor.
#4y^2(y+2)-1(y+2)=0#
#(y+2)(4y^2-1)=0#

Recall that the zero-product principle tells, when #ab=0#, then either #a=0# or #b=0#, or both #a# and #b# are #0#.

#(y+2)(4y^2-1)=0#
#y+2=0 or 4y^2-1=0#
#y=-2 or y^2=1/4#
#y=-2 or y=1/2 or y=-1/2#

Reminder: #sqrt (1/4)=+-1/2# , there will be 2 answers for square root

Here is the answer :)
Hope it can help you.