Question

find all the zeroes of the polynomial x^3-4x^2-3x+12,if two of its zeroes are ⅓×3 and ⅓×-3​

Answer 1

*Step by step explanation:-

Given polynomial is:-

• x³ — 3x² — 4x + 12

x³ — 3x² — 4x + 12Substituting x = 2 in the given

polynomial, we get,

• polynomial, we get,(2)³ — 3(2)² — 4(2) + 12

• polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12

• polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12= 0

polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12= 0So x = 2 is a zero of the given polynomial.

polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12= 0So x = 2 is a zero of the given polynomial.Substituting x = 3 in the given polynomial, we get,

• polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12= 0So x = 2 is a zero of the given polynomial.Substituting x = 3 in the given polynomial, we get,(3)³ — 3(3)² — 4(3) + 12

• polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12= 0So x = 2 is a zero of the given polynomial.Substituting x = 3 in the given polynomial, we get,(3)³ — 3(3)² — 4(3) + 12= 27 — 27 — 12 + 12

• polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12= 0So x = 2 is a zero of the given polynomial.Substituting x = 3 in the given polynomial, we get,(3)³ — 3(3)² — 4(3) + 12= 27 — 27 — 12 + 12= 0

polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12= 0So x = 2 is a zero of the given polynomial.Substituting x = 3 in the given polynomial, we get,(3)³ — 3(3)² — 4(3) + 12= 27 — 27 — 12 + 12= 0So x = 3 is a zero of the given polynomial.

polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12= 0So x = 2 is a zero of the given polynomial.Substituting x = 3 in the given polynomial, we get,(3)³ — 3(3)² — 4(3) + 12= 27 — 27 — 12 + 12= 0So x = 3 is a zero of the given polynomial.Since x = 2 and x = 3 are the zeros of the polynomial so (x—2) and (x—3) will be the factors of the polynomial.

polynomial, we get,(2)³ — 3(2)² — 4(2) + 12= 8 — 12 — 8 + 12= 0So x = 2 is a zero of the given polynomial.Substituting x = 3 in the given polynomial, we get,(3)³ — 3(3)² — 4(3) + 12= 27 — 27 — 12 + 12= 0So x = 3 is a zero of the given polynomial.Since x = 2 and x = 3 are the zeros of the polynomial so (x—2) and (x—3) will be the factors of the polynomial.Let third zero be a, so we can write:-

• (x—2)(x—3)(x—a) = x³ — 3x² — 4x + 12

• (x—2)(x—3)(x—a) = x³ — 3x² — 4x + 12=> (x² — 5x + 6)(x—a) = x³ — 3x² — 4x + 12

• (x—2)(x—3)(x—a) = x³ — 3x² — 4x + 12=> (x² — 5x + 6)(x—a) = x³ — 3x² — 4x + 12=> x³ - 5x² + 6x - ax² + 5ax - 6a = x³ - 3x² - 4x + 12

(x—2)(x—3)(x—a) = x³ — 3x² — 4x + 12=> (x² — 5x + 6)(x—a) = x³ — 3x² — 4x + 12=> x³ - 5x² + 6x - ax² + 5ax - 6a = x³ - 3x² - 4x + 12Comparing the constant terms, we get:-

• —6a = 12

• a = -12/6

• => a = —2

=> a = —2So the third zero is —2.

$hope \: it \: helps \: you$