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Unit 05: Polynomials @math-11-nbf:sol
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ts will be able to * State and prove remainder theorem and explain through examples. * Find remainder ... zeros of a polynomial. * State and prove factor theorem. * Use factor theorem to factorize a cubic polynomial. * Apply concepts of remainder and factor theorem to real world problems. <panel type="default
Question 1, Exercise 5.1 @math-11-nbf:sol:unit05
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1(i)===== Find the remainder by using 'Remainder Theorem': $2 x^{3}+3 x^{2}-4 x+1$ is divided by $x+2$. *... }-4 x+1$\\ $x-c=x+2 \implies c=-2$. By Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c)... 1(ii)===== Find the remainder by using 'Remainder Theorem': $x^{4}+2 x^{3}-x^{2}+2 x+3$ is divided by $x-2$... - c = x - 2 \implies c = 2 \). By the Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c)
Question 1 and 2, Exercise 5.2 @math-11-nbf:sol:unit05
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=====Question 1===== Factorize by using factor theorem: $y^{3}-7 y-6$ ** Solution. ** Suppose $f(y)=y^... -7 (-1)-6 \\ &= -1+7-6 =0. \end{align*} By factor theorem, $y+1$ is factor of $f(y)$. Using synthetic div... =====Question 2===== Factorize by using factor theorem: $2 x^{3}-x^{2}-2 x+1$ ** Solution. ** \begin{a... &= 2 - 1 - 2 + 1 = 0. \end{align*} By the factor theorem, \( x - 1 \) is a factor of \( f(x) \). Using sy
Question 3 and 4, Exercise 5.2 @math-11-nbf:sol:unit05
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=====Question 3===== Factorize by using factor theorem: $2 x^{3}+5 x^{2}-9 x-18$ ** Solution. ** Suppo... 6 + 20 + 18 - 18 = 0. \end{align*} By the factor theorem, \( x + 2 \) is a factor of \( f(x) \). Using sy... $ =====Question 4===== Factorize by using factor theorem: $3 x^{3}-5 x^{2}-36$ **Solution.** Suppose \( ... &= 81 - 45 - 36 = 0. \end{align*} By the factor theorem, \( x - 3 \) is a factor of \( f(x) \). Using sy
Question 5 and 6, Exercise 5.2 @math-11-nbf:sol:unit05
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=====Question 5===== Factorize by using factor theorem: $t^{3}+t^{2}+3 t-5$ ** Solution. ** Suppose \(... + 1 + 3 - 5 \\ &= 0. \end{align*} By the factor theorem, \( t - 1 \) is a factor of \( f(t) \). Using sy... actors. **Solution.** It is given by the factor theorem, \( x - 2 \) is a factor of \( f(x) \).Then Usin
Question 2 and 3, Exercise 5.1 @math-11-nbf:sol:unit05
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}-5 x+6$ and $x-c=x-3$ $\implies c=3$. By factor theorem $x-3$ is factor of $p(x)$ iff $p(3)=0$. Now \be... }-5 x+1$ and $x-c=x-3$ $\implies c=3$. By factor theorem $x-3$ is factor of $p(x)$ iff $p(3)=0$. Now \be
Question 4 and 5, Exercise 5.1 @math-11-nbf:sol:unit05
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-7 x+6$ and $x-c=x+1$ $\implies c=-1$. By factor theorem $x+1$ is factor of $p(x)$ iff $p(-1)=0$. This gi
Question 6 and 7, Exercise 5.1 @math-11-nbf:sol:unit05
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and $x-c=x-2$ $\implies c=2$. \\ By the Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c)
Question 4 & 5, Review Exercise @math-11-nbf:sol:unit05
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frac{0}{9} = 0. \end{align*} Hence by the factor theorem, \( 3y - 2 \) is a factor of \( 6y^{3} - y^{2} -
Question 6 & 7, Review Exercise @math-11-nbf:sol:unit05
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\) by \( (x - 4) \) is zero. \\ By the remainder theorem, the remainder is \( p(4) \). Since \( p(4) = 0
Question 14, Exercise 8.1 @math-11-nbf:sol:unit08
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er wire makes with the ground, then by Pythagoras theorem \begin{align*} & \overline{AD}^2 = \overline{AB}^