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# help

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42
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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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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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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)

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
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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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Its fine, mistakes happen...

Vinculum  Apr 14, 2022
#10
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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

Apr 14, 2022