How do I solve these???? (1 Viewer)

[nimesha11]

1. Solve the equation 2x^4 + 9x^3 - 26x^2 - 36x +72 = 0 for x given that the sum of two of its roots is zero

2. Given that the roots of the polynomial p(x) = ax^3 + bx^2 +cx +d form a geometric sequence, show that ac^3 = b^3d

3. When a polynomial is divided by (x-a), the remainder is a^3. When the same polynomial is divided by (x-b), the remainder is b^3. Find the remainder when this polynomial is divided by (x-a)(x-b).

 

[Drongoski]

Q1) P(x) = (x-2)(x+2)(2x-3)(x+6) = 0

.: 4 roots are -6, -2, 2 and 1.5

How did I figure this out? Well I first tried the expected conventional approach of letting the 4 roots to be: -a, a, b and c and setting up the system of equations involving the coeffs of terms in P(x). I found, on first attempt, that the equations were relatively messy to solve; maybe I did not look at things the right way. So I changed tack.

I did the trial-&-error approach of seeking zeros of P(x), trying in succession x = 1, -1, 2 .... I found P(2) = 0. So I tried out x = -2; (to determine if 2 & -2 are the pair of roots) again I got 0. So P(x) has (x-2) and (x+2) , ( therefore x² - 4) as factors. By inspection or by long division, I found P(x) = (x² - 4)(2x² + 9x - 18). You can now solve for the 2nd quadratic eqn.

 

[ichila101]

For Q2)

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[ichila101]

Q3)

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[nimesha11]

Q3)

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But how did you know the remainder was in the form of ax+b?

 

[nimesha11]

Thank you so much

 

[ichila101]

But how did you know the remainder was in the form of ax+b?

The remainder is either equal to 0 (in which case it would be a factor) or it is a degree less than the divisor. Since the divisor was a quadratic then the remainder would be linear in the form ax+b

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[nimesha11]

ok got it Thank you so much!