Section 5.4 : Finding Zeroes of Polynomials
3. Find all the zeroes of the following polynomial.
A(x)=2x4−7x3−2x2+28x−24Show All Steps Hide All Steps
Start SolutionWe’ll need all the factors of -24 and 2.
−24:±1,±2,±3,±4,±6,±8,±12,±242:±1,±2 Show Step 2Here is a list of all possible rational zeroes for the polynomial.
±1±1=±1±2±1=±2±3±1=±3±4±1=±4±6±1=±6±8±1=±8±12±1=±12±24±1=±24±1±2=±1±2±2±2=±1±3±2=±3±2±4±2=±2±6±2=±3±8±2=±4±12±2=±6±24±2=±12So, we have a total of 20 possible zeroes for the polynomial.
Show Step 3We now need to start the synthetic division work. We’ll start with the “small” integers first.
2−7−228−24−12−9721−45=A(−1)≠012−5−721−3=A(1)≠0−22−1120−120=A(−2)=0!!Okay we now know that x=−2 is a zero and we can write the polynomial as,
A(x)=2x4−7x3−2x2+28x−24=(x+2)(2x3−11x2+20x−12) Show Step 4So, now we need to continue the process using Q(x)=2x3−11x2+20x−12. Here is a list of all possible zeroes of Q(x).
±1±1=±1±2±1=±2±3±1=±3±4±1=±4±6±1=±6±12±1=±12±1±2=±1±2±2±2=±1±3±2=±3±2±4±2=±2±6±2=±3±12±2=±6So we have a list of 16 possible zeroes, but note that we’ve already proved that ±1 can’t be zeroes of the original polynomial and so can’t be zeroes of Q(x) either.
Here’s the synthetic division work for this Q(x).
2−1120−12−22−1550−112=Q(−2)≠022−760=Q(2)=0!!Therefore, x=2 is a zero of Q(x) and the factored form of Q(x) is,
Q(x)=2x3−11x2+20x−12=(x−2)(2x2−7x+6)This also means that the factored form of the original polynomial is now,
A(x)=2x4−7x3−2x2+28x−24=(x+2)(x−2)(2x2−7x+6) Show Step 5We’re down to a quadratic polynomial and so we can and we can just factor this to get the fully factored form of the original polynomial. This is,
A(x)=2x4−7x3−2x2+28x−24=(x+2)(x−2)(x−2)(2x−3)=(x+2)(x−2)2(2x−3) Show Step 6From the fully factored form we get the following set of zeroes for the original polynomial.
\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 2\hspace{0.25in}\hspace{0.25in}x = \frac{3}{2}\hspace{0.25in}\hspace{0.25in}x = 2\,\,\,({\mbox{multiplicity 2}})}}