If a polynomial function f(x) has roots 0, 4, and 3+ /11, what must also be a root of f(x)?
1 answer:
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Answer:
2x + 1
Step-by-step explanation:
Factor out 2x² - 5x-3
divide x - 3 by x - 3
=> 2x + 1
<em><u>Or another step</u></em>
Step by Step Solution
STEP
1
:
Equation at the end of step 1
STEP
2
:
2x² - 5x - 3/ x - 3
Trying to factor by splitting the middle term
2.1 Factoring 2x² - 5x - 3
The first term is, 2x² its coefficient is 2 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is -3
Step-1 : Multiply the coefficient of the first term by the constant 2 • -3 = -6
Step-2 : Find two factors of -6 whose sum equals the coefficient of the middle term, which is -5 .
-6 + 1 = -5 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -6 and 1
2x² - 6x + 1x - 3
Step-4 : Add up the first 2 terms, pulling out like factors :
2x • (x-3)
Add up the last 2 terms, pulling out common factors :
1 • (x-3)
Step-5 : Add up the four terms of step 4 :
(2x+1) • (x-3)
Which is the desired factorization
Canceling Out :
2.2 Cancel out (x-3) which appears on both sides of the fraction line.
Final result :
2x + 1
