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11. Solve: (x - 2)(x ^ 2 - 9x + 13) + 5 = 0​

User Areg
by
7.8k points

2 Answers

8 votes
x
3
−9x
2
+13x−2x
2
+18x−26+5=0
User Meghann
by
8.5k points
9 votes

Answer:


x=1, x=3, x=7

Explanation:

Start multiplying and adding all together. You get
x^3-11x^2+31x-21 =0.

At this point we're stuck factoring since there exist a formula for the roots, but it's nowhere easy to remember. We start writing down The List (rational root theorem: it's the divisor of the constant term divided by the divisor of the lead coefficent, with a plus or minus, in our case it's
\pm1, \pm3, \pm7, \pm 21) and let's try checking one by one.

Value
+1 gives a zero on the LHS, it means the polynomial is divisible by
(x-1). By doing the division, or writing it as
(x-1)(ax^2+bx+c) = x^3-11x^2+31x-21, multiplying on the LHS and then finding the values of a, b, and c (easiest for me, your mileage may vary) we can write our expression, finally, as


(x-1)(x^2-10x+21) =0 At this point you can either apply the quadratic formula or decompose again the second factor as


(x-3)(x-7) by finding two numbers whose sum is -10 and product is 21, ie -3 and -7. At this point the expression


(x-1)(x-3)(x-7)=0 is true if
x=1, x=3, x=7

User John Chadwick
by
8.1k points

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