Question

find the zero of the polynomial

1) 8x^3-6x^2-x+2
2) 3x^2-2

Answer 1

Answer:

Brainly.in

Question

find the zeroes of the polynomial x^2+1/6x-2 and verify the realtion between the coefficient and zeroes of the polynomial.

Answer · 95 votes

Given: The polynomial x^2 + 1/6(x) - 2 To find: The zeroes of the polynomial and verify the relation between the coefficient and zeroes of the polynomial.Solution:Now we have given the polynomial: x^2 + 1/6(x) - 2 = 0Simplifying it, we get: 6x^2 + x - 12 = 0So by splitting middle term, we get: 6x^2 - 8x + 9x - 12 = 0Now combining the terms, we get: (6x^2 - 8x) + (9x - 12) = 0 2x(3x - 4) + 3(3x + 4) = 0 (3x - 4)(2x + 3) = 0 x = 4/3 or x = -3/2So verifying it, we get:Sum of zeroes: 4/3 - 3/2 = -1/6Product of zeroes: 4/3 x (-3/2) = -2Hence verified.Answer: So the zeroes of the polynomial are 4/3 and -3/2.

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Brainly.in

Question

Find the zeroes of quadratic polynomial p(x)=6x^2+18x and verify relationships between zeroes and coefficients

Answer · 7 votes

=6x²-7x-3=6x²+2x-9x-3=2x(3x+1)-3(3x+1)=(2x-3)(3x+1)⇒2x-3=0 or ⇒3x+1=0⇒x=3/2 or ⇒ x= - 1/3α=3/2 ,β= - 1/3⇒α+β= -b/a⇒3/2+(-1/3)= - (-3)/6⇒3/2-1/3=1/2⇒7/6 =1/2 ⇒αβ=c/a ⇒ 3/2(-1/3)= -7/6⇒-1/2= -7/6 ⇒1/2=7/6 #Tanwar

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Sarthaks eConnect

Question

Find the zeroes of the polynomials: (i) x^3 + 6x^2 + 11x + 6

Answer · 1 vote

(i) x3 + 6x2 + 11x + 6 Let the given polynomial be p(x) = x3 + 6x2 + 11x + 6 The coefficient of the leading term is 1 and the constant term is 6. Also the of 6 are 1, 2 and 3. So, the possible integral zeroes of p(x) are ± 1, ± 2, and ±3. Now, p(x) does not have a negative coefficient of any term. So, p(x) cannot be zero for positive integral value of x. Again Hence, the integral zeroes of the given polynomial are -1, -2 and -3. (ii) x3 + 2x2 – x – 2 The coefficient of the leading term is 1 and the constant is - 2. Also the factors of 2 are 1 and 2. So, the possible integral zeroes of p(x) are ±1, ±2. (iii) x4 – 2x3 – 7x2 + 8x + 12 (iv) x3 – 2x3 – x + 2 Let p(x) = x3 – 2x3 – x + 2 The coefficient of the leading term is 1 and the constant term is 2. Also the factors of 2 are 1 and 2. So the possible integral zeros of p(x) are ± 1 and ±2. (v) x3 – 3x2 – 9x – 5 Let p(x) = x3 – 3x2 – 9x – 5 The coefficient of the leading term is 1 and the constant term is -5. Also f…

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Toppr

Question

The zeros of polynomial p(x) = x^2 - 3x + 2 are

Answer · 16 votes

The zeros of polynomial p(x) = x^2 - 3x + 2 can be given y p(x) = 0 x^2 - 3x + 2 = 0 x^2 - 2x - x + 2 = 0 x(x - 2) - 1(x - 2) = 0 (x - 1)(x - 2) = 0 x = 1,2

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Toppr

Question

If one of the zero of polynomial 3x^2 - 8x + 2k + 1 is seven times the other, find the value of k.

Answer · 22 votes

Let the zero of polynom

Step-by-step explanation:

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