Question

CLA SS 10 th rs Agarwal maths question .....DONT MAKE SPAMS BRAINLIEST YR ans.awarded. plz explain in detail that how we take minus common in this solution​

Answer 1

Refer attachment

Given that 1 is a zero of given polynomial, hence (x - 1) would be a factor of given polynomial.

Dividing (7x - x³ - 6) by (x - 1) gives ( - x² - x + 6) (Refer attachment)

Now, the roots of equation ( - x² - x + 6) would be the roots of given cubic polynomial (7x - x³ - 6) too, as

(x - 1)(-x² - x + 6) = 7x - x³ - 6

So, now finding the roots of - x² - x + 6

→ - x² - x + 6 = 0

→ x² + x - 6 = 0

→ x² + 3x - 2x - 6 = 0

→ x(x + 3) - 2(x + 3) = 0

→ (x - 2)(x + 3) = 0

Hence,

(x - 1)(-x² - x + 6) = 7x - x³ - 6

→ (x - 1)(x - 2)(x + 3) = 7x - x³ - 6

Hence, the roots of (7x - x³ - 6) are 1, 2 and (- 3)

Answer 2

Given that:

• 1 is zero of (7x-x^3-6).

To Find:

• The other zero.

SOLUTION:

Since 1 is zero of f(x).

So, x-1 can be taken out as a factor from f(x).

f(x) = (x-1)(-x^2-x+6)

Now, factorising (-x^2-x+6), we get

(-x^2-x+6)= 0

-x^2-3x+2x+6 = 0

-x(x+3)+2(x+3) = 0

(x+3)(2-x) = 0

x = -3 and x = 2

So, the other zeros are -3 and 2.

Additional Information:

• A cubic polynomial has at max. 3 roots.

• Sum of roots = -b/a.

• Sum of roots taken product of two at a time = c/a

• Product of roots = -d/a

where,

a is coefficient of x^3

b is coefficient of x^2

c is coefficient of x

d is constant term.

Hope this helps