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g(x)=x^3+2x^2-17x+6

1)show that (x-3) is a factor of g(x)

2)hence express g(x) as a product of a linear factor and a quadratic factor

3)hence solve the equation g(x)=0 leaving any answers in an exact form

 

 

please can you put the number of the question before the answer as it is less confusing

thank u

 Apr 14, 2022
 #1
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+1

3)

 

g(x) = 0 when x = 3, (-sqrt(33)-5)/2, (sqrt(33)-5)/2

 Apr 14, 2022
 #4
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thanks can u answer 1 and 2 as it got me all over the place.

Guest Apr 14, 2022
 #2
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-3

Answer this one if ur so smart Mr potato aka 8 year older

 Apr 14, 2022
 #3
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done

ch1ck3n  Apr 14, 2022
 #6
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maybe you are smart but y u had to tel me say that in my past on the other question

Kakashi  Apr 14, 2022
 #5
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+1

1)

 

Applying the quadratic forumla gives us (x-3)(x^2+5x-2)
 

2)

 

(x-3)(x^2+5x-2)

 Apr 14, 2022
edited by ch1ck3n  Apr 14, 2022
 #7
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are u shor that question 1 and 2 are the same

Guest Apr 14, 2022
 #8
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Question 1 asks if x-3 is in the equation, which it does

 

Quesition 2 asks for the full expanded form

ch1ck3n  Apr 14, 2022
 #9
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thanks

Guest Apr 14, 2022
 #11
avatar+578 
+1

I don't think you can apply the quadratic formula to get question one.

Vinculum  Apr 14, 2022
 #12
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oh c**p didnt realise

ch1ck3n  Apr 14, 2022
 #13
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+2

Its fine, mistakes happen...

 

smileysmileysmiley

Vinculum  Apr 14, 2022
 #10
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+3

1 and 2)

 

Using the Rational root theorem

 

\(\left(x-3\right)\frac{x^3+2x^2-17x+6}{x-3}\)

 

\(\frac{x^3+2x^2-17x+6}{x-3} = x^2+5x-2\)

 

\(\left(x-3\right)\left(x^2+5x-2\right)\)

 

 

 

3)

 \(0^3+0x^2-17\cdot \:0+6=6\)

 

-Vinculum

smileysmileysmiley

 Apr 14, 2022

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