Question

If (x+2) is a factor of p(x)=ax^3+bx^2+x-6 and p(x) when divided by(x-2) leaves the remainder 4. Prove that a=0 and b=2 Please answer ASAP..i am getting graded on this

Answer 1

Let p(x) = ax³ + bx² + x - 6

Let p(x) = ax³ + bx² + x - 6A/C to question,

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0⇒-8a + 4b - 8 = 0

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0⇒-8a + 4b - 8 = 0⇒ 2a - b + 2 = 0 -------------(1)

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0⇒-8a + 4b - 8 = 0⇒ 2a - b + 2 = 0 -------------(1)again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0⇒-8a + 4b - 8 = 0⇒ 2a - b + 2 = 0 -------------(1)again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.so, P(2) = a(2)³ + b(2)² + 2 - 6 = 4

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0⇒-8a + 4b - 8 = 0⇒ 2a - b + 2 = 0 -------------(1)again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.so, P(2) = a(2)³ + b(2)² + 2 - 6 = 4⇒8a + 4b - 4 = 4

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0⇒-8a + 4b - 8 = 0⇒ 2a - b + 2 = 0 -------------(1)again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.so, P(2) = a(2)³ + b(2)² + 2 - 6 = 4⇒8a + 4b - 4 = 42a + b -2 = 0 -------------(2)

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0⇒-8a + 4b - 8 = 0⇒ 2a - b + 2 = 0 -------------(1)again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.so, P(2) = a(2)³ + b(2)² + 2 - 6 = 4⇒8a + 4b - 4 = 42a + b -2 = 0 -------------(2)solve equations (1) and (2),

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0⇒-8a + 4b - 8 = 0⇒ 2a - b + 2 = 0 -------------(1)again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.so, P(2) = a(2)³ + b(2)² + 2 - 6 = 4⇒8a + 4b - 4 = 42a + b -2 = 0 -------------(2)solve equations (1) and (2),4a = 0 ⇒a = 0 and b = 2

Let p(x) = ax³ + bx² + x - 6A/C to question,(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0⇒-8a + 4b - 8 = 0⇒ 2a - b + 2 = 0 -------------(1)again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.so, P(2) = a(2)³ + b(2)² + 2 - 6 = 4⇒8a + 4b - 4 = 42a + b -2 = 0 -------------(2)solve equations (1) and (2),4a = 0 ⇒a = 0 and b = 2Then, equation will be 2x² + x - 6

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