Question

2. Let P (x) = 2x4 +15x3 +31x2 +20x +4

(a) Determine whether (x −1) is a factor of P (x). (2)

(b) Find all the possible rational zeros of P (x) by using the Rational Zeros Theorem. (2)

(c) Solve P (x) = 0

Answer 1

2.

(a)

$$P(1)=2(1)^4 +15(1)^3 +31(1)^2 +20(1)+4$$

$$=72\not=0$$

Therefore $(x −1)$ is not a factor of $P (x).$

(b) Since all coefficients are integers, we can apply the Rational Zeros Theorem.

The coefficient of the constant term is 4. Find its factors: $\pm1, \pm2, \pm4.$

These are the possible values for $p.$

The leading coefficient is $2.$ Find its factors: $\pm1, \pm2.$

These are the possible values for $q.$

Find all possible values of $p/q:$ $\pm\dfrac{1}{2}, \pm1, \pm2, \pm4.$

Possible rational roots: $\pm\dfrac{1}{2}, \pm1, \pm2, \pm4.$

(c)

Check the possible roots

$$\dfrac{1}{2}: P(\dfrac{1}{2})=2(\dfrac{1}{2})^4 +15(\dfrac{1}{2})^3 +31(\dfrac{1}{2})^2 +20(\dfrac{1}{2})+4$$

$$=\dfrac{95}{4}\not=0$$

$$-\dfrac{1}{2}: P(-\dfrac{1}{2})=2(-\dfrac{1}{2})^4 +15(-\dfrac{1}{2})^3 +31(-\dfrac{1}{2})^2$$

$$+20(-\dfrac{1}{2})+4=0$$

$$1: P(1)=2(1)^4 +15(1)^3 +31(1)^2 +20(1)+4$$

$$=72\not=0$$

$$-1: P(-1)=2(-1)^4 +15(-1)^3 +31(-1)^2$$

$$+20(-1)+4=2\not=0$$

$$2: P(2)=2(2)^4 +15(2)^3 +31(2)^2 +20(2)+4$$

$$=320\not=0$$

$$-2: P(-2)=2(-2)^4 +15(-2)^3 +31(-2)^2$$

$$+20(-2)+4=0$$

$$4: P(4)=2(4)^4 +15(4)^3 +31(4)^2 +20(4)+4$$

$$=2052\not=0$$

$$-4: P(-4)=2(-4)^4 +15(-4)^3 +31(-4)^2$$

$$+20(-4)+4=-28\not=0$$

Actual rational roots: $-\dfrac{1}{2}, -2.$

$$2x^4+15x^3 +31x^2 +20x +4$$

$$=(2x+1)(x+2)(x^2+5x+2)$$

$$x^2+5x+2=0$$

$$D=25-8=17$$

$$x=\dfrac{-5\pm\sqrt{17}}{2}$$

Total roots:$-\dfrac{1}{2}, -2,\dfrac{-5-\sqrt{17}}{2},\dfrac{-5+\sqrt{17}}{2}$