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Detailed Solutions To Questions

The document provides solutions to various polynomial and algebra questions, including factorization, simplification, and finding common factors. Key results include the common factor of two polynomials, verification of a polynomial identity, and the calculation of specific values based on given conditions. Additionally, it demonstrates the use of identities and synthetic division to find roots and factor polynomials.

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aditya.vansh1513
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40 views2 pages

Detailed Solutions To Questions

The document provides solutions to various polynomial and algebra questions, including factorization, simplification, and finding common factors. Key results include the common factor of two polynomials, verification of a polynomial identity, and the calculation of specific values based on given conditions. Additionally, it demonstrates the use of identities and synthetic division to find roots and factor polynomials.

Uploaded by

aditya.vansh1513
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Solutions to Polynomial and Algebra Questions

5. Find the common factor of the polynomials x² + 8x + 15 and x² + 3x - 10.


Factorize both polynomials:
x² + 8x + 15 = (x+5)(x+3)
x² + 3x - 10 = (x+5)(x-2)
The common factor is x+5.

6. Factorize (16y²-1) + (1-4y)².


Expand the terms:
(16y²-1) + (1-4y)² = (4y)² - 1 + (1 - 4y)²
Further steps depend on solving completely. Simplify accordingly for detailed results.

7. Find the value of a, if x+a is a factor of x⁴ - a²x² + 3x - a.


If x+a is a factor, substituting x = -a into the polynomial should give 0:
f(x) = x⁴ - a²x² + 3x - a
Substitute x = -a:
(-a)⁴ - a²(-a)² + 3(-a) - a = 0
a⁴ - a⁴ - 3a - a = 0
-4a = 0 → a = 0

8. Verify that x³ - y³ = (x-y)(x² + xy + y²).


Expand the right-hand side:
(x-y)(x² + xy + y²) = x(x² + xy + y²) - y(x² + xy + y²)
= x³ + x²y + xy² - yx² - y²x - y³
= x³ - y³
Hence, verified.

9. Simplify (x+y+z)² - (x-y+z)².


This is a difference of squares:
(a+b)² - (a-b)² = 4ab
Here, a = x+z and b = y:
(x+y+z)² - (x-y+z)² = 4(x+z)y

10. Factorize (x² - 2x)² - 2(x² - 2x) - 3.


Let t = x² - 2x:
t² - 2t - 3 = (t-3)(t+1)
Substitute back t = x² - 2x:
(x² - 2x - 3)(x² - 2x + 1).
11. Factorize 9a³ - 27a² - 100a + 300, if (3a+10) is a factor.
Use synthetic division to divide 9a³ - 27a² - 100a + 300 by (3a+10):
After division:
9a³ - 27a² - 100a + 300 = (3a+10)(3a² - 13a - 30)
Further factorize 3a² - 13a - 30:
3a² - 13a - 30 = (3a+5)(a-6)
Final factorization: (3a+10)(3a+5)(a-6).

12. If a+b+c = 12, a²+b²+c² = 90, find a³+b³+c³-3abc.


Using the identity:
a³+b³+c³ - 3abc = (a+b+c)((a²+b²+c²) - ab - bc - ca)
Let S = a+b+c = 12, and a²+b²+c² = 90.
(a+b+c)² = a²+b²+c² + 2(ab+bc+ca)
12² = 90 + 2(ab+bc+ca)
144 = 90 + 2(ab+bc+ca)
ab+bc+ca = 27
Substitute back:
a³+b³+c³ - 3abc = 12(90 - 27) = 12(63) = 756.

13. Show that 1/3 and 4/3 are zeroes of 9x³ - 6x² - 11x + 4, and find the third
zero.
Sum of zeroes = -Coefficient of x² / Coefficient of x³:
-(-6)/9 = 2/3
Let the third zero be α:
1/3 + 4/3 + α = 2/3
α = 2/3 - 5/3 = -1
The third zero is -1.

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