Math, asked by dfpdanceforpassion, 3 months ago

The polynomial p(x) = ax3 + 9x2 + 4z - 8, when divided by (x + 3) leaves -20
as remainder. Find the value of a.​

Answers

Answered by vanshgupta12345
0

Answer:

x+3=0

x= -3

By remainder theorem:-

p [-3] = a(-3)^3 +9 (-3)^2 +4 (-3) -8

-20 = -27a +81 -12 -8

-20 = -27a +81 -20

-81 = -27a

a = 3

Answered by vipinkumar212003
0

Answer:

\blue{\mathfrak{\underline{\large{Given}}}:} p(x) = ax^3 + 9x^2  +  4x - 8.  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\ \\  \blue{\mathfrak{\underline{\large{Given}}}:} g(x) = x  +  3.  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \\  \\  \blue{\mathfrak{\underline{\large{By \: the \: remainder \: theorem, \: we \: get}}}:}  \\ = > x+3 = 0 \\  = > x = -3.  \\ \blue{\mathfrak{\underline{\large{plug \: x = -3 \: in \: p(x), \: we \: get}}}:}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \\  \\ = > a( - 3)^3 + 9( - 3)^2  +  4( - 3) - 8 = -20 \\  =>  - 27a + 81   -  12 - 8 = -20  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \\  =>  - 27a + 61 = -20  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  =>  - 27a = -81  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \\  = >a= 3. \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \\ \\  \blue{\mathfrak{\large{Therefore \: the \: value \: of \: a = 3.}}}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\ \red{\mathfrak{\underline{\large{Hope \: It \: Helps \: You }}}} \\ \green{\mathfrak{\underline{\large{Mark \: Me \: Brainliest}}}}

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