if (x+1)and (x-3)are two factors of x³-ax²-bx+3 find a and b hence find the third factor
Answers
Step-by-step explanation:
Let p(x) = x3 + ax2 + bx +6
(x-2) is a factor of the polynomial x3 + ax2 + b x +6
p(2) = 0
p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0
7 +2 a +b = 0
b = - 7 -2a -(i)
x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.
p(3) = 3
p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3
11+3a +b =1 => 3a+b =-10 => b= -10-3a -(ii)
Equating the value of b from (ii) and (i) , we have
(- 7 -2a) = (-10 - 3a)
a = -3
Substituting a = -3 in (i), we get
b = - 7 -2(-3) = -7 + 6 = -1
Thus the values of a and b are -3 and -1 respectively.
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EXPLANATION.
If ( x + 1 ) and ( x - 3 ) are the factor of the polynomial,
⇒ x³ - ax² - bx + 3 = 0.
To find the value of a and b.
Case = 1.
⇒ x + 1 = 0.
⇒ x = -1.
put the value of x = -1 in equation we get,
⇒ (-1)³ - a(-1)² - b(-1) + 3 = 0.
⇒ -1 - a + b + 3 = 0.
⇒ -a + b + 2 = 0.
⇒ a = b + 2 ......(1).
Case = 2.
⇒ x - 3 = 0.
⇒ x = 3.
put the value of x = 3 in equation we get,
⇒ (3)³ - a(3)² - b(3) + 3 = 0.
⇒ 27 - 9a - 3b + 3 = 0.
⇒ -9a - 3b + 30 = 0. ........(2).
put the value of equation (1) in equation (2) we get,
⇒ -9 ( b + 2 ) - 3b + 30 = 0.
⇒ -9b - 18 - 3b + 30 = 0.
⇒ -12b + 12 = 0.
⇒ -12b = -12
⇒ b = 1.
put the value of b = 1 in equation (1) we get,
⇒ a = b + 2.
⇒ a = 1 + 2.
⇒ a = 3.
Value of a = 3 and b = 1.
MORE INFORMATION.
(1) = A quadratic equation whose roots are α,β.
sum of zeroes of quadratic equation
α + β = -b/a.
product of zeroes of quadratic equation.
αβ = c/a.
(2) = A cubic equation whose zeroes are α,β,γ.
sum of zeroes of cubic equation.
α + β + γ = -b/a.
product of zeroes of cubic equation.
αβγ = -d/a.
product of zeroes of cubic two at a times.
αβ + βγ + γα = c/a.
(3) = A biquadratic equation whose zeroes are α,β,γ,δ.
σ₁ = α + β + γ + δ = -b/a.
σ₂ = αβ + αγ + αδ + βγ + βδ + γδ = c/a.
σ₃ = αβγ + αβδ + αγδ + βγδ = -d/a.
σ₄ = αβγδ = e/a.