SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given polynomials

One of the zeroes is:

STEP 2: Use synthetic division to divide the polynomials

Write the coefficients of the numerator


![\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column \\ 1\left(-5\right)=-5 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}]()

![\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column: \\ 1\left(-5\right)=-5 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}]()

![\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column: \\ \left(-20\right)\left(-5\right)=100 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:100}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:\:}\end{matrix} \end{gathered}]()


The last carry-down value is the remainder and it is 0 (zero)
Since the remainder is a zero, hence, x=-5 is a zero
Step 3: Answer question b
To get the factors, the remainder of the division in step 2 is given as:
The remaining factors of f(x) is:

STEP 4: Answer Question c

The complete factorization will give:

STEP 5: Answer question d
The zeroes of f(x) will be:

zeroes are: -5,4,-5
STEP 6: Plot the graph