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# help

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164
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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
+1

3)

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

Apr 14, 2022
#4
0

thanks can u answer 1 and 2 as it got me all over the place.

Guest Apr 14, 2022
#2
-3

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

Apr 14, 2022
#3
0

done

ch1ck3n  Apr 14, 2022
#6
-2

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
+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
0

are u shor that question 1 and 2 are the same

Guest Apr 14, 2022
#8
0

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
#11
+1

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

Vinculum  Apr 14, 2022
#12
0

oh c**p didnt realise

ch1ck3n  Apr 14, 2022
#13
+2

Its fine, mistakes happen...   Vinculum  Apr 14, 2022
#10
+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   Apr 14, 2022