Question

factorise x^3 - 13x + 12 = ​

Answer 1

Answer:

Ur answer:

Step-by-step explanation:

Step-by-step explanation:If a polynomial function has integer coefficients, then every rational zero will have the form

p/q

, where p is a factor of the constant and q is a factor of the leading coefficient.

, where p is a factor of the constant and q is a factor of the leading coefficient.Here

, where p is a factor of the constant and q is a factor of the leading coefficient.Herep=±1,±2,±3,±4,±6,±12 and

, where p is a factor of the constant and q is a factor of the leading coefficient.Herep=±1,±2,±3,±4,±6,±12 and q=±1

, where p is a factor of the constant and q is a factor of the leading coefficient.Herep=±1,±2,±3,±4,±6,±12 and q=±1Find every combination of ±

p/q

.

. These are the possible roots of the polynomial function.

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we get

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1)

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1)

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1)

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1) =(−1) 3−13(−1)12

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1) =(−1) 3−13(−1)12=−1+13−12

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1) =(−1) 3−13(−1)12=−1+13−12=13+13

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1) =(−1) 3−13(−1)12=−1+13−12=13+13=0

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1) =(−1) 3−13(−1)12=−1+13−12=13+13=0

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1) =(−1) 3−13(−1)12=−1+13−12=13+13=0

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1) =(−1) 3−13(−1)12=−1+13−12=13+13=0 ∴(x+1) is a factor of f(x)

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1) =(−1) 3−13(−1)12=−1+13−12=13+13=0 ∴(x+1) is a factor of f(x) Now, dividing f(x) by (x+1), we get

. These are the possible roots of the polynomial function.±1,±2,±3,±4,±6,±12Substituting x=1 in f(x), we getf(−1) =(−1) 3−13(−1)12=−1+13−12=13+13=0 ∴(x+1) is a factor of f(x) Now, dividing f(x) by (x+1), we getx =3−13x−12

=(x+1)(x 2 −x−12)

=(x+1)(x 2 −x−12)=(x+1)(x 2 −4x+3x−12)

=(x+1)(x 2 −x−12)=(x+1)(x 2 −4x+3x−12)=(x+1){x(x−4))+3(x−4)}

=(x+1)(x 2 −x−12)=(x+1)(x 2 −4x+3x−12)=(x+1){x(x−4))+3(x−4)}=(x+1)(x+3)(x−4)

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