Polynomial Functions and Linear Inequalities

POLYNOMIAL FUNCTIONS AND LINEAR INEQUALITIES PLOTS

Finding Zeros of Polynomial Functions

Finding the rational zeros of fx= x3+2×2-11x-12 Given that all the coefficients are integers, we employ the Rational Zeros Theorem.

The Coefficient of the constant term is -12

Finding its factors: ±1, ±2, ±3, ±4, ±6, ±12 – These represents the possible values for p.

The leading coefficient (Coefficient of the term with the highest degree) is 1.

Finding its factors: ±1 – These are the possible values for q. Finding all possible values of pq: ±11,±21,±31,±41,±61,±121Simplifying and removing duplicates, the possible Rational Roots: ±1, ±2, ±3, ±4, ±6, ±12Checking the Possible Roots:

If a is a root of the polynomialP(x), the remainder from the division of Pxby x-a should be equal to

Checking 1: Divide x3+2×2-11x-12 by x-1The quotient is x3+3x-8 and the remainder is -20.

Checking -1: Divide x3+2×2-11x-12 by x+1The quotient is x2+x-12 and the remainder is

Therefore -1 is a root.

Checking 2: Divide x3+2×2-11x-12 by x-2The quotient is x2+4x-3 and the remainder is -18.

Checking -2: Divide x3+2×2-11x-12 by x+2The quotient is x2-11 and the remainder is 10Checking 3: Divide x3+2×2-11x-12 by x-3The quotient is x2+5x+4 and the remainder is 0.

Thus, 3 is a root.

Checking -3: Divide x3+2×2-11x-12 by x+3The quotient is x2-x-8 and the remainder is 12.

Checking 4: Divide x3+2×2-11x-12 by x-4The quotient is x2+6x+13 and the remainder is 40.

Checking -4: Divide x3+2×2-11x-12 by x+4The quotient is x2-2x-3 and the remainder is

Therefore, -4 is a root.

Checking 6: Divide x3+2×2-11x-12 by x-6The quotient is x2+8x+37 and the remainder is 210.

Checking -6: Divide x3+2×2-11x-12 by x+6The quotient is x2-4x+13 and the remainder is -90Checking 12: Divide x3+2×2-11x-12 by x-12The quotient is x2+14x+157 and the remainder is 1872.

Checking -12: Divide x3+2×2-11x-12 by x+12The quotient is x2-10x+109 and the remainder is -1320

ANSWER

Possible Rational Roots: ±1, ±2, ±3, ±4, ±6, ±12Actual Rational Roots: -1, 3, -4Finding the rational zeros of fx= 2×3-x2-41x-20 Given that all the coefficients are integers, we employ the Rational Zeros Theorem.

The Coefficient of the constant term (trailing coefficient) is -20

Finding its factors: ±1, ±2, ±4, ±5, ±10, ±20 – These represents the possible values for p.

The leading coefficient (Coefficient of the term with the highest degree) is 2.

Finding its factors: ±1, ±2 – These are the possible values for q. Finding all possible values of pq: ±11,±12,±21,±22,±41,±42, ±51,±52,±101,±102,±201,±202Simplifying and removing duplicates, the possible Rational Roots:

±1, ±12,±2, ±4, ±5, ±52±10, ±20Checking the Possible Roots:

If a is a root of the polynomialP(x), the remainder from the division of Pxby x-a should be equal to

Checking 1: Divide 2×3-x2-41x-20 by x-1The quotient is 2×2+x-40 and the remainder is -60.

Checking -1: Divide 2×3-x2-41x-20 by x+1The quotient is 2×2-3x-38 and the remainder is 18Checking 12: Divide 2×3-x2-41x-20 by x-12The quotient is 2×2-41 and the remainder is -812.

Checking -12: Divide 2×3-x2-41x-20 by x+12The quotient is 2×2-2x-40 and the remainder is

Therefore -12 is a root.

Checking 2: Divide 2×3-x2-41x-20 by x-2The quotient is 2×2+3x-35 and the remainder is -90.

Checking -2: Divide 2×3-x2-41x-20 by x+2The quotient is 2×2-5x-31 and the remainder is 42Checking 4: Divide 2×3-x2-41x-20 by x-4The quotient is 2×2+7x-13 and the remainder is -72

Checking -4: Divide 2×3-x2-41x-20 by x+4The quotient is 2×2+7x-13 and the remainder is -72.

Checking 5: Divide 2×3-x2-41x-20 by x-5The quotient is 2×2+9x+4 and the remainder is 0.

Therefore, 5 is a root.

Checking -5: Divide 2×3-x2-41x-20 by x+5The quotient is 2×2-11x+14 and the remainder is -90

Checking52: Divide 2×3-x2-41x-20 by x-52The quotient is 2×2+4x-31 and the remainder is -1952.

Checking -52: Divide 2×3-x2-41x-20 by x+52The quotient is 2×2-6x-26 and the remainder is 45Checking 10: Divide 2×3-x2-41x-20 by x-10The quotient is 2×2+19x+149 and the remainder is 1470.

Checking -10: Divide 2×3-x2-41x-20 by x+10The quotient is 2×2-21x+169 and the remainder is -1710Checking 20: Divide 2×3-x2-41x-20 by x-20The quotient is 2×2+39x+739 and the remainder is 14760.

Checking -20: Divide 2×3-x2-41x-20 by x+20The quotient is 2×2-41x+779 and the remainder is -15600

ANSWER

Possible Rational Roots: ±1, ±12,±2, ±4, ±5, ±52±10, ±20Actual Rational Roots: -12, -4, 5Finding the rational zeros of fx= x4+x3-5×2+x-6 Given that all the coefficients are integers, we employ the Rational Zeros Theorem.

The Coefficient of the constant term is -6

Finding its factors: ±1, ±2, ±3, ±6 – These represents the possible values for p.

The leading coefficient (Coefficient of the term with the highest degree) is 1.

Finding its factors: ±1 – These are the possible values for q. Finding all possible values of pq: ±11,±21,±31,±61Simplifying and removing duplicates, the possible Rational Roots: ±1, ±2, ±3, ±6Checking the Possible Roots:

If a is a root of the polynomialP(x), the remainder from the division of Pxby x-a should be equal to

Checking 1: Divide x4+x3-5×2+x-6 by x-1The quotient is x3+2×2-3x-2 and the remainder is -8.

Checking -1: Divide x4+x3-5×2+x-6 by x+1The quotient is x3+2×2-3x-2 and the remainder is -12

Checking 2: Divide x4+x3-5×2+x-6 by x-2The quotient is x3+3×2+x+3 and the remainder is 0

Therefore 2 is a root.

Checking -2: Divide x4+x3-5×2+x-6 by x+2The quotient is x3-x2-3x+ and the remainder is -20Checking 3: Divide x4+x3-5×2+x-6 by x-3The quotient is x3+4×2+7x+22 and the remainder is 60.

Checking -3: Divide x4+x3-5×2+x-6 by x+3The quotient is x3-2×2+x-2 and the remainder is 0.

Therefore -3 is a root.

Checking 6: Divide x4+x3-5×2+x-6 by x-6The quotient is x3+7×2+37x+223 and the remainder is 1332.

Checking -6: Divide x4+x3-5×2+x-6 by x+6The quotient is x3-5×2+25x-149 and the remainder is 888

ANSWER

Possible Rational Roots: ±1, ±2, ±3, ±6Actual Rational Roots: 2,- 3Linear Inequalities Plots

(a).

y≤5x>-2

(b).

x+y≥6y<2x+8

(c)

x≥0y≥03x-4y<12