Search
You can find the results of your search below.
Fulltext results:
- Question 6 and 7, Exercise 5.1
- Solution. ** Suppose $p(x)=x^3-7x+6$.\\ $1$ will be zero of $p(x)$ if $p(1)=0$. Thus \begin{align*} p... $1$ is the zero of p(x). \\ Similarly, $-2$ will be zero of $p(x)$ if f $p(-2)=0$. \begin{align*} p(-
- Question 1, Review Exercise
- }-2 x^{2}+5$ is divided by $x+1$, then $x+1$ will be its:\\ * (a) divisor as well as factor\\ ... $x^{3}+5 x^{2}-4 x+k$, then the value of $k$ will be:\\ * (a) $-4$ * (b) $-20$ * %%(c)%%
- Question 2, Exercise 5.3
- , the number of tickets sold during the match can be modeled by $t(x)=x^{3}-12 x^{2}+48 x+74$, where $
- Question 4 & 5, Review Exercise
- al. ** Solution. ** Let the required polynomial be \( f(x) \). Given the zeros \( 4, \frac{3}{5}, -2
- Question 6 & 7, Review Exercise
- \\ &= 48 + k. \end{align*} For the remainder to be zero: \[ 48 + k = 0. \] Thus, \[ k = -48. \]