690 Worked-Out Solutions to Exercises: Chapters 21 to 29
We found the real rootsx=−5/3 or x= 7/4Let’s put −5/3 in place of x, and then carry out the calculations. We get[3× (−5/3)+ 5][16 × (−5/3)^2 − 56 × (−5/3)+ 49] = 0The first term in square brackets is equal to 0. Let’s check:3 × (−5/3)+ 5
=−15/3+ 5
=− 5 + 5
= 0This means the entire expression must be 0; it doesn’t matter what the second term is.
Now let’s insert 7/4 for x. We have only to work with the trinomial term this time, and it
comes out equal to 0. Let’s try it:16 × (7/4)^2 − 56 × (7/4) + 49
= 16 × 49/16 − 392/4 + 49
= 49 − 98 + 49
=− 49 + 49
= 0The whole expression must equal 0 because the second factor is 0.- We begin by setting up the synthetic division array with the “test root,” x= 5, and the
coefficients in the top row, like this:
5 − 9 21 104 80
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####Subsequent steps proceed as follows.5 − 9 21 104 80
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− 9 ###5 − 9 21 104 80
− 45 ##
− 9 ###