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- Question 3, Exercise 5.3
- condition \begin{align*} & x(2x)(2x+2) = 144 \\ \implies & 4x^2(x+1)=144 \\ \implies & x^2(x+1)=36 \\ \implies & x^3+x^2-36=0 \end{align*} Suppose p(x)=x3+x2−36. Since \begin{align*}
- Question 4, Exercise 5.3
- ndition \begin{align*} & x(2x+3)(x-2) = 2475 \\ \implies & x(2x^2+3x-4x-6)=2475 \\ \implies & x(2x^2-x-6)-2475=0 \\ \implies & 2x^3-x^2-6x-2475=0 \end{align*} Suppose p(x)=2x3−x2−6x−2475. Sin
- Question 5, Exercise 5.3
- ow \begin{align*} & Length \times Width = Area\\ \implies & Length \times (2x+8) = 6 x^{2}+38 x+56 \\ \implies & Length \times (2x+8) = (2x+8)(3x+7) \\ \implies & Length = 3x+7 \\ \end{align*} Hence length of rectang
- Question 1, Exercise 5.1
- Given: p(x)=2x3+3x2−4x+1\\ $x-c=x+2 \implies c=-2$. By Remainder Theorem, we have \begin{alig... + 2x^{3} - x^{2} + 2x + 3 \) \\ \( x - c = x - 2 \implies c = 2 \). By the Remainder Theorem, we have \beg
- Question 2 and 3, Exercise 5.1
- * Let p(x)=x3−2x2−5x+6 and x−c=x−3 $\implies c=3.Byfactortheoremx-3isfactorofp(x)... * Let p(x)=x3−2x2−5x+1 and x−c=x−3 $\implies c=3.Byfactortheoremx-3isfactorofp(x)
- Question 1, Exercise 5.3
- on, we have \begin{align*} & x(x+3)(x+10)=120 \\ \implies & x(x^2+3x+10x+30)-120=0\\ \implies & x^3+13x^2+30x-120=0. \end{align*} Consider $$p(x)=x^3+13x^2+30x-12
- Question 4 and 5, Exercise 5.1
- * Let p(x)=x3+qx2−7x+6 and x−c=x+1 $\implies c=-1.Byfactortheoremx+1isfactorofp(x
- Question 6 and 7, Exercise 5.1
- Let p(x)=2x3+3x2−3x−m and x−c=x−2 $\implies c=2$. \\ By the Remainder Theorem, we have \begin
- Question 8 and 9, Exercise 5.1
- +3)(2x+1) \end{align*} i.e. x+3=0 or 2x+1=0 $\impliesx=-3orx=-\dfrac{1}{2}.Hence2,3$ and
- Question 10, Exercise 5.1
- olume of room = area of floor × height. $\implies x^{3}+11 x^{2}+34 x+24=(x^2+10x+24)(x+1)$ Hence