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- Question 4 & 5, Review Exercise
- on. ** Given \begin{align*}3y-2&=0\\ 3y&=2\\ y&=\frac{2}{3}\end{align*} Suppose \begin{align*} f(y) &= 6y^{3} - y^{2} - 5y + 2\\ f\left(\frac{2}{3}\right) &= 6\left(\frac{2}{3}\right)^{3} - \left(\frac{2}{3}\right)^{2} - 5\left(\frac{2}{3}\right) + 2 \\ &= 6\left(\frac{8}{2
- Question 7 and 8, Exercise 5.2
- ==== Factorize $2 x^{3}-15 x^{2}+27 x-10$ if ' $\dfrac{1}{2}$ ' is one of its zero. ** Solution. ** Us... ynthetic division to divide \( f(x) \) by \( x - \frac{1}{2} \): \begin{align} \begin{array}{r|rrrr} \frac{1}{2} & 2 & -15 & 27 & -10 \\ & & 1 & -7 & 10 ... n} This gives: \begin{align*} f(x) &= \left(x - \frac{1}{2}\right)(2x^{2} - 14x + 20)\\ &=\left(x - \fr
- Question 2 & 3, Review Exercise
- (4 y-2) \quad$ ** Solution. ** \begin{align*} \frac{(64 y^{3}-8)}{(4 y-2)}&= \frac{(4y - 2)(16y^{2} + 8y + 4)}{4y - 2}\\ & = 16y^{2} + 8y + 4 .\end{align*}... t) \div(5 y-2)$ ** Solution. ** \begin{align*} \frac{(125 y^{3}-8)}{(5 y-2)} &= \frac{(5y - 2)\left(25y^{2} + 10y + 4\right)}{5y - 2} \\ & = 25y^{2} + 10y +
- Question 1, Review Exercise
- "a2" collapsed="true">(a): $ 0$</collapse> iii. $\frac{x^{2}-x-9}{x-3}=x+2+\frac{?}{x-3}$\\ * (a) $-27$\\ * (b)$-3$\\ * %%(c)%% $\frac{3}{x-3}+x+2$\\ * (d) $ 3$ \\ <btn type="link
- Question 8 and 9, Exercise 5.1
- e. $x+3=0$ or $2x+1=0$ $\implies$ $x=-3$ or $x=-\dfrac{1}{2}$. Hence $2$, $3$ and $-\dfrac{1}{2}$ are the roots of given polynomial. GOOD =====Question 9=
- Question 5, Exercise 5.3
- $3x+4$ Now Length of side of square $ABFG$ = $\dfrac{1}{2}(2 x+8)$ = $x+4$ Area of square $ABFG$ = (x
- Question 6 & 7, Review Exercise
- iding one polynomial by another is $3 x^{2}-x+32-\frac{121}{x+4}$. What is the dividend? ** Solution. *
- Question 7, Review Exercise
- iding one polynomial by another is $3 x^{2}-x+32-\frac{121}{x+4}$. What is the dividend? ** Solution. *