factorise x cube minus x square + 14 x plus 24
Answers
Answer:
P(x) = x³ - x² - 14 x + 24
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist.
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ]
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ] then compare coefficients of x² term or x term, then
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ] then compare coefficients of x² term or x term, then
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ] then compare coefficients of x² term or x term, then -a x² - 2 x² = - 1 x² => a = -1
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ] then compare coefficients of x² term or x term, then -a x² - 2 x² = - 1 x² => a = -1 2 a x -12 x = - 14 x => a = -1
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ] then compare coefficients of x² term or x term, then -a x² - 2 x² = - 1 x² => a = -1 2 a x -12 x = - 14 x => a = -1 P(x) = ( x -2 ) ( x² +x-12) =
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ] then compare coefficients of x² term or x term, then -a x² - 2 x² = - 1 x² => a = -1 2 a x -12 x = - 14 x => a = -1 P(x) = ( x -2 ) ( x² +x-12) =
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ] then compare coefficients of x² term or x term, then -a x² - 2 x² = - 1 x² => a = -1 2 a x -12 x = - 14 x => a = -1 P(x) = ( x -2 ) ( x² +x-12) = find factors of -12 that have difference of -1.
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ] then compare coefficients of x² term or x term, then -a x² - 2 x² = - 1 x² => a = -1 2 a x -12 x = - 14 x => a = -1 P(x) = ( x -2 ) ( x² +x-12) = find factors of -12 that have difference of -1. so -4 and 3 are roots of x² +x -12
P(x) = x³ - x² - 14 x + 24 factors of 24 are possible rational roots of P(x), if at all rational roots exist. so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24 or -1, -2, -3, -4, -6, -8, -12, -24 check with 1, -1, 2 .. x = 2 => P(2) = 0 let, P(x) = (x - 2) [ x² - a x + (24/-2) ] then compare coefficients of x² term or x term, then -a x² - 2 x² = - 1 x² => a = -1 2 a x -12 x = - 14 x => a = -1 P(x) = ( x -2 ) ( x² +x-12) = find factors of -12 that have difference of -1. so -4 and 3 are roots of x² +x -12 P(x) = (x - 2 ) ( x + 4 ) ( x - 3)