Question 7 and 8, Exercise 5.2

Solutions of Question 7 and 8 of Exercise 5.2 of Unit 05: Polynomials. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Factorize $2 x^{3}-15 x^{2}+27 x-10$ if ' $\dfrac{1}{2}$ ' is one of its zero.

Solution.

Using synthetic 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 \\ \hline & 2 & -14 & 20 & 0 \\ \end{array} \end{align}

This gives: \begin{align*} f(x) &= \left(x - \frac{1}{2}\right)(2x^{2} - 14x + 20)\\ &=\left(x - \frac{1}{2}\right)(2x^{2} - 14x + 20) \\ &= \left(x - \frac{1}{2}\right)(2x^{2} - 10x - 4x + 20 )\\ &= \left(x - \frac{1}{2}\right)[2x(x - 5) - 4(x - 5) ]\\ &=\frac{1}{2}\left(2x - 1\right)2 (x - 2)(x - 5)\\ &=\left(2x - 1\right) (x - 2)(x - 5) \end{align*}

Finally, the complete factorization is: \begin{align*} f(x) &= \left(2x - 1\right) (x - 2)(x - 5). \end{align*}

If $h(x)=4 x^{3}+4 x^{2}+73 x+36$ and $h\left(\frac{-1}{2}\right)=0$, then factorize $h(x)$.

Solution.

Given: $h(x)=4 x^{3}+4 x^{2}+73 x+36$ and $h\left(\frac{-1}{2}\right)=0$.

This given $-\frac{1}{2}$ is zeor of $h(x)$. By using synthetic division: \begin{align} \begin{array}{r|rrrr} -\frac{1}{2} & 4 & 4 & 73 & 36 \\ & \downarrow & -2 & -1 & -36 \\ \hline & 4 & 2 & 72 & 0 \\ \end{array}\end{align}

Thus \begin{align*} h(x) &= \left(x+\frac{1}{2} \right)(4x^2+2x+72) \\ &= \left(\frac{2x+1}{2} \right)(2)(2x^2+x+36) \\ &= \left(2x+1\right)(2x^2+x+36). \end{align*}

< Question 5 & 6