Answer to Question #339505 in Calculus for Momo

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