Question

The polynomial p(y) is obtained by
subtracting -y2 - 6y – 6 from y3 - y2 + y +6.
How many terms are there in p(y)?
Number of terms =​

Answer 1

Here,

• On subtracting $\rm{ { - y}^{2} - 6y - 6}$ from $\rm {y}^{3} - {y}^{2} + y + 6$, p(y) has been obtained.

To find :- Number of terms in p(y)

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

$\huge \mathbb{SOLUTION : }$

• To get p(y), we need to subtract $\rm{ { - y}^{2} - 6y - 6}$ from $\rm {y}^{3} - {y}^{2} + y + 6$.

$\tiny = > \rm {y}^{3} - {y}^{2} + y + 6 - (\rm{ { - y}^{2} - 6y - 6} ) \\ \tiny = > \rm {y}^{3} - {y}^{2} + y + 6 { {y}^{2} + 6y + 6} \\ \tiny{ = > \underline{ \: \: \rm{ {y}^{3} + {5y}^{2} + 7y + 6} \: \: }}$

Thus,

• We can state that p(y) has four (4) terms.