Question

Please answer Question no.38 and 40

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Answer 1

38.

$\begin{aligned}&a^3+\dfrac{3ax}{8}+\dfrac{x^3}{64}-\dfrac{1}{8}\\ \\ \Longrightarrow\ \ &a^3+\frac{x^3}{64}-\dfrac{1}{8}+\dfrac{3ax}{8}\\ \\ \Longrightarrow\ \ &(a)^3+\left(\frac{x}{4}\right)^3+\left(-\dfrac{1}{2}\right)^3-3\bigg(a\bigg)\bigg(\frac{x}{4}\bigg)\left(-\dfrac{1}{2}\right)\end{aligned}$

Now it seems in the form  p³ + q³ + r³ - 3pqr.

We know,   $p^3+q^3+r^3-3pqr=(p+q+r)(p^2+q^2+r^2-pq-qr-rp).$  

So,

$\displaystyle(a)^3+\left(\frac{x}{4}\right)^3+\left(-\dfrac{1}{2}\right)^3-3\bigg(a\bigg)\bigg(\frac{x}{4}\bigg)\left(-\dfrac{1}{2}\right)\\ \\ \\ \\ =\left(a+\frac{x}{4}-\dfrac{1}{2}\right)\left(a^2+\frac{x^2}{16}+\frac{1}{4}-\frac{ax}{4}+\dfrac{x}{8}+\dfrac{a}{2}\right)$

Hence, factorized!

40.  Note the mistake that the LCM is actually (x - 2)(x - 1)(x + 2)²(x + 3).

Consider the polynomial  (x² + x - 2)(x² + x - a).

    (x² + x - 2)(x² + x - a)

⇒  (x^2 - x + 2x - 2)(x² + x - a)

⇒  (x(x - 1) + 2(x - 1))(x² + x - a)

⇒  (x - 1)(x + 2)(x² + x - a)

Now, compare this polynomial with the LCM  (x - 2)(x - 1)(x + 2)²(x + 3).  We can find that  (x - 1)(x + 2)  is common in both, so  (x - 2)(x + 2)(x + 3)  is a multiple of  (x² + x - a).

With an assumption of  x² + x - a = (x - 2)(x + 3) = x² + x - 6  [because coefficients of x in both are equal],  we get,

a = 6

{Since both  (x + 2)(x + 3)  and  (x² + x - a) are polynomials with same degree, there makes a contradiction at the coefficient of x being different; one is 5 while the other is 1. That's why LCM given in the question is changed and (x - 2) is multiplied with it. If  (x - 1)  was multiplied with the given LCM instead of (x - 2), an error would occur at  (x² + x - b)(x² + 5x + a).}

Now, consider the polynomial  (x² + x - b)(x² + 5x + a).

By substituting  a = 6,  we get  (x² + x - b)(x² + 5x + 6).

    (x² + x - b)(x² + 5x + 6)

⇒  (x² + x - b)(x² + 2x + 3x + 6)

⇒  (x² + x - b)(x(x + 2) + 3(x + 2))

⇒  (x² + x - b)(x + 2)(x + 3)

Now compare this with the LCM  (x - 2)(x - 1)(x + 2)²(x + 3).  We see that (x + 2)(x + 3) is common in both, so that  (x - 2)(x - 1)(x + 2)  is a multiple of (x² + x - b).

With an assumption of  x² + x - b = (x - 1)(x + 2) = x² + x - 2  [because coefficients of x in both are equal],  we get,

b = 2

Hence found!