# Determine for which polynomials (x+2) is a factor. Explain your answer. P(x)=x^4-3x^3-16x-12 Q(x)=x^3-3x^2-16x-12

$$\displaystyle{P}{\left({x}\right)}={x}^{{4}}-{3}{x}^{{3}}-{16}{x}-{12}$$

$$\displaystyle{Q}{\left({x}\right)}={x}^{{3}}-{3}{x}^{{2}}-{16}{x}-{12}$$

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averes8

We have to show if (x+2) is a factor of the polynomials $$\displaystyle{P}{\left({x}\right)}={x}^{{4}}-{3}{x}^{{3}}-{16}{x}-{12}$$ and $$\displaystyle{Q}{\left({x}\right)}={x}^{{3}}-{3}{x}^{{2}}-{16}{x}-{12}$$ or not.
To check the factor we have to see whether (x+2) satisfies the given polynomials.
If it satisfies the given polynomial or gives the answer 0 i.e P(x)=0 or Q(x)=0 we will say that it is a factor of the polynomial, if it doesn't satisfy the given polynomial we will say that it is not a factor of the polynomial.
First, we will put $$\displaystyle{x}+{2}={0}\Rightarrow{x}=-{2}$$
Now, we will substitute x=-2 in the given polynomials P(x) and Q(x) and see if it is a factor of these polynomials or not.
Now,
$$\displaystyle{P}{\left(-{2}\right)}={\left(-{2}\right)}^{{4}}-{3}{\left(-{2}\right)}^{{3}}-{16}{\left(-{2}\right)}-{12}$$
$$\displaystyle{P}{\left(-{2}\right)}={16}+{24}+{32}-{12}$$
$$\displaystyle{P}{\left(-{2}\right)}={60}$$
therefore, $$\displaystyle{P}{\left(-{2}\right)}\ne{0}$$
hence, $$x=-2$$ is not a factor of the polynomial P(x)
Now,
$$\displaystyle{Q}{\left(-{2}\right)}={\left(-{2}\right)}^{{3}}-{3}{\left(-{2}\right)}^{{2}}-{16}{\left(-{2}\right)}-{12}$$
$$\displaystyle{Q}{\left(-{2}\right)}=-{8}-{12}+{32}-{12}$$
$$\displaystyle{Q}{\left(-{2}\right)}={0}$$
therefore, $$Q(-2)=0$$
Hence, $$x=-2$$ is a factor of the polynomial $$Q(x)$$