Victor Wall

2021-12-16

All the real zeros of the given polynomial are integers. Find the zeros, and write the polynomial in factored form.
$P\left(x\right)={x}^{4}-2{x}^{3}-3{x}^{2}+8x-4$

Ben Owens

Step 1
To find zeroes of polynomial
$P\left(x\right)={x}^{4}-2{x}^{3}-3{x}^{2}+8x-4$
Step 2
P(1)=0
so
${x}^{4}-2{x}^{3}-3{x}^{2}+8x-4=\left(x-1\right)\left({x}^{3}-{x}^{2}-4x+4\right)$
Also x=1 is zero of $\left({x}^{3}-{x}^{2}-4x+4\right)$
Again
${x}^{4}-2{x}^{3}-3{x}^{2}+8x-4=\left(x-1\right)\left({x}^{3}-{x}^{2}-4x+4\right)$
$={\left(x-1\right)}^{2}\left({x}^{2}-4\right)$
$±2$ is zero of $\left({x}^{2}-4\right)$
${x}^{4}-2{x}^{3}-3{x}^{2}+8x-4=\left(x-1\right)\left(x-1\right)\left(x-2\right)\left(x+2\right)$

Joseph Lewis

The given polynomial is $P\left(x\right)={x}^{4}-2{x}^{3}-3{x}^{2}+8x-4$
Since the leading term is 1, any rational zero must be a divisior of the constant term 4
So the possible rational zeros are $±1,±2,±4$
We test each of these possibilities
$P\left(1\right)={1}^{4}-2{\left(1\right)}^{3}-3{\left(1\right)}^{2}+8\left(1\right)-4=1-2-3+8-4=0$
$P\left(-1\right)={\left(-1\right)}^{4}-2{\left(-1\right)}^{3}-3{\left(-1\right)}^{2}+8-\left(1\right)-4=1+2-3-8-4=-12$
$P\left(2\right)={2}^{4}-2{\left(2\right)}^{3}-3{\left(2\right)}^{2}+8-\left(2\right)-4=16-16-12+16-4=0$
$P\left(-2\right)={\left(-2\right)}^{4}-2{\left(-2\right)}^{3}-3{\left(-2\right)}^{2}+8-\left(-2\right)-4=16+16-12-16-4=0$
$P\left(4\right)={4}^{4}-2{\left(4\right)}^{3}-3{\left(4\right)}^{2}+8\left(4\right)-4=256-128-48+32-4=108$
$P\left(-4\right)={\left(-4\right)}^{4}-2{\left(-4\right)}^{3}-3{\left(-4\right)}^{2}+8\left(-4\right)-4=-256+128-48-32-4=300$
The rational zeros of P are 1,2, and -2.

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