What can be the solution of question-19
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vol:
(x-1)
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Step 1 :
Equation at the end of step 1 :
(((2 • (x3)) + 5x2) - 5x) - 2
Step 2 :
Equation at the end of step 2 :
((2x3 + 5x2) - 5x) - 2
Step 3 :
Checking for a perfect cube :
3.1 2x3+5x2-5x-2 is not a perfect cube
Trying to factor by pulling out :
3.2 Factoring: 2x3+5x2-5x-2
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -5x-2
Group 2: 5x2+2x3
Pull out from each group separately :
Group 1: (5x+2) • (-1)
Group 2: (2x+5) • (x2)
Equation at the end of step 1 :
(((2 • (x3)) + 5x2) - 5x) - 2
Step 2 :
Equation at the end of step 2 :
((2x3 + 5x2) - 5x) - 2
Step 3 :
Checking for a perfect cube :
3.1 2x3+5x2-5x-2 is not a perfect cube
Trying to factor by pulling out :
3.2 Factoring: 2x3+5x2-5x-2
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -5x-2
Group 2: 5x2+2x3
Pull out from each group separately :
Group 1: (5x+2) • (-1)
Group 2: (2x+5) • (x2)
Attachments:
but what is the integral zero
if an integer a is a zero of a polynomial function withintegral. coefficients and a leading coefficient of 1, then a is a factor of the constant term of the. polynomial.Integral Zeros
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