please solve question number 17 to 24 do it in your notebook and then click the picture and then post it here then I'll make him or her brain list
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here is your answer
17) x(x-2)(x+3)
we can write x as x+o
(x+0)(x-2)(x+3) = 0
x+0 = 0
x= 0
x-2 = 0
x = 0+2 = 2
x+3 = 0
x = 0-3 = -3
zeroes are 0, 2 and -3.
hope his helps you
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19)8x^3 + √27y^3
(2x)^3 + (√3y)^3
using identity a^3+b^3 =
(a+b)(a^2-ab+b^2)
= (2x+√3y)[(2x)^2-2x×√3y+(√3y)^2]
(2x+√3y)(4x^2-2√3xy+3y^2)
plz mark as brainliest answer plzzzzz
17) x(x-2)(x+3)
we can write x as x+o
(x+0)(x-2)(x+3) = 0
x+0 = 0
x= 0
x-2 = 0
x = 0+2 = 2
x+3 = 0
x = 0-3 = -3
zeroes are 0, 2 and -3.
hope his helps you
plz mark as brainliest answer plzzzzz
19)8x^3 + √27y^3
(2x)^3 + (√3y)^3
using identity a^3+b^3 =
(a+b)(a^2-ab+b^2)
= (2x+√3y)[(2x)^2-2x×√3y+(√3y)^2]
(2x+√3y)(4x^2-2√3xy+3y^2)
plz mark as brainliest answer plzzzzz
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