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- Question 10, Exercise 5.1
- ====== Question 10, Exercise 5.1 ====== Solutions of Question 10 of Exercise 5.1 of Unit 05: Polynomials. This is unit o... tbook Board, Islamabad, Pakistan. =====Question 10===== A rectangular room has a volume of $\left(x^... } -1 & 1 & 11 & 34 & 24 \\ & \downarrow & -1 & -10 & -24 \\ \hline & 1 & 10 & 24 & 0 \\ \end{array}
- Question 7 and 8, Exercise 5.2
- =Question 7===== Factorize $2 x^{3}-15 x^{2}+27 x-10$ if ' $\dfrac{1}{2}$ ' is one of its zero. ** So... egin{array}{r|rrrr} \frac{1}{2} & 2 & -15 & 27 & -10 \\ & & 1 & -7 & 10 \\ \hline & 2 & -14 & 20 & 0 \\ \end{array} \end{align} This gives: \begin{al... + 20) \\ &= \left(x - \frac{1}{2}\right)(2x^{2} - 10x - 4x + 20 )\\ &= \left(x - \frac{1}{2}\right)[2x
- Question 1, Exercise 5.3
- f bottle = $x+3$ cm\\ Height of bottle = $x+3+7=x+10$ cm\\ Volume of bottle = $120 cm^3$ By given condtion, 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 $... dth of bottle = $2+3=5cm$\\ Height of bottle = $2+10=12 cm$\\ GOOD ====Go to ==== <text align=
- Question 4 and 5, Exercise 5.1
- istan. =====Question 4===== If $4 y^{3}-4 y^{2}+10+2 y$ is completely divisible by any of its factor such that the quotient is $4 y^{2}-8 y+10$, then find other factor. ** Solution. **
- Question 2 & 3, Review Exercise
- {3}-8)}{(5 y-2)} &= \frac{(5y - 2)\left(25y^{2} + 10y + 4\right)}{5y - 2} \\ & = 25y^{2} + 10y + 4. \end{align*} ====Go to ==== <text align="lef
- Question 4 & 5, Review Exercise
- ht) + 2 \\ &= \frac{48}{27} - \frac{4}{9} - \frac{10}{3} + 2 \\ &= \frac{16}{9} - \frac{4}{9} - \frac{... - 23x + 12 )(x + 2)\\ &= \frac{1}{5}\left(5x^3 + 10x^2 - 23x^2 - 46x + 12x + 24\right) \\ &= \frac{1}
- Question 8 and 9, Exercise 5.1
- text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5-1-p6|Question 10 >]]</btn></text>
- Question 1, Review Exercise
- : $(x+1)(x-2)$</collapse> ii. Divide $9 y^{2}+9 y-10$ by $3 y-2$, then remainder is:\\ * (a) $ 0$\