Page 16 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1.39 If the polynomial 6x − 5xy − 4y − 11x + 22y + m can be factored into the
2
2
product of two linear polynomials, find the value of m and factor the polynomial.

2
2
2
2
Solution: Let 6x − 5xy − 4y − 11x + 22y + m = (2x + y + k)(3x − 4y + l) =6x − 5xy − 4y +
2
2
2
2
6x − 5xy − 4y − 11x + 22y + m = (2x + y + k)(3x − 4y + l) =6x − 5xy − 4y +
(3k +2l)x +(l − 4k)y + kl
(3k +2l)x +(l − 4k)y + kl, then equaling the coefficients:
3k +2l = −11,l − 4k = 22, kl = m, which result in k = −5,l =2,m = −10.
2 2 2 2
2 2
2 2
2 2
2 2
Hence, 6x − 5xy − 4y − 11x + 22y + m =6x − 5xy − 4y − 11x + 22y − 10 = (2x + y −
6x − 5xy − 4y − 11x + 22y + m =6x − 5xy − 4y − 11x + 22y − 10 = (2x + y
6x − 5xy − 4y − 11x + 22y + m =6x − 5xy − 4y − 11x + 22y − 10 = (2x + y − −
5)(3x − 4y + 2)
5)(3x − 4y + 2) 2).
5)(3x − 4y +
1.40 Given |x +4| + |3 − x| = 10 −|y − 2|− |1+ y| , find the maximum and minimum
values of xy .
Solution: |x +4| + |3 − x| = 10 −|y − 2|−|1+ y|⇒ |x +4| + |3 − x| + |y − 2| + |1+ y| = 10 .
Since |x +4| + |3 − x|≥ 7 and |y − 2| + |1+ y|≥ 3. |x +4| + |3 − x| + |y − 2| + |1+ y| = 10
only if we choose equal sign in both inequalities.
|x +4| + |3 − x|≥ 7 ⇒−4 ≤ x ≤ 3;

|y − 2| + |1+ y|≥ 3 ⇒−1 ≤ y ≤ 2.

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