Help me with this please.

Question

find m so that x-3 is a factor of 7x^3-23x^2+mx+12

Answer 1

x - 3 is a factor of a polynomial function f(x) only when f(3) = 0.

So, 7(3)³ - 23(3)² + 3m + 12 = 0

Solve for m.

Answer 2

So John Y.,

Just start by writing out the terms of the big polynomial (the "dividend"), with a little lateral space between them!, and the divisor (x-3) to the left, in the standard format of a "long division". You will be carrying out a division, just the same way as you (probably) used to do with whole numbers, BUT with the difference that successive "positions" represent decrementing powers of x (and their coefficients) in a polynomial, NOT decrementing powers of ten in an integer.

So, what is your first "position" of the quotient? Compare the coefficients of the leading term of the dividend and divisor: 7 and (implicitly) 1 respectively. How many times does 1 go into 7? 7, right? So, above the "7x^3" term, write the start of the quotient: (7x^3)/x = 7x^2 . Next, just as you would for a long division problem, multiply that first term of the quotient (7x^2) through the entire divisor (x-3), and write the result under the respective terms of the dividend, that will be (7x^3 - 21x^2) . Just as you would in a long division problem, subtract this entire expression from the respective places of the dividend above it. Note that the second bit (-21x^2) is subtracted, so it is equivalent to adding 21x^2 to the -23x^2 term immediately above it. I suggest that you write the expression to be subtracted in parentheses, with a minus sign in front of it, so that you are constantly tipped off about that situation! Write the result of the subtraction under the respective terms, etc. You have now removed all the x^3 (zeroed them out in the division), and have -2x^2 as the result. Bring down the next term of the dividend (+mx) and continue the long division process. How many times does 1x go into -2x^2? -2x, right? So write that in the quotient line, it represents a -2x term in the quotient. As you did before, multiply that term (-2x) through the entire divisor (x-3), and write the product under the respective columns of your problem. You should have written (-2x^2+6x). Subtract that entire expression as before, you should be left with no x^2 term, and (m-6)x (after you bring down the next term and add it in to the -6x). Now you are ready for the last division term of the problem -- but, oh my!, you have that lead coefficient (m-6) staring at you! So, take a little sidestep and consider that last term of the dividend: (+12). Reason backwards: if this division is going to come out with NO REMAINDER, which is what you want (otherwise, (x-3) could not be a factor of the polynomial), you will need to subtract the constant "12" in the next division step. But a "12" could only come from the product of -3 in the divisor and some other constant value, which you immediately know MUST be -4, right? So, multiply the -4 by the (x-3), and write the expression which must be subtracted in the last "step" ; it must be (-4x+12). Look twice at that to make sure that you haven't made any mistakes in subtraction, etc.!

OK, now you know what the coefficient subtracted in the x column position is, it is -4. That's the equivalent of adding +4. So when 4 is added to (m-6), that must come out to zero (to remove the x terms from the remainder under process). So (m-6+4) = 0 . I think you can take it from there?

I hope this clues you in on how to do a polynomial "long division" factoring. You just did it with explicitly showing the powers of x; if your course hasn't yet gotten to it, you will undoubtedly learn how to do all this without writing out as text the powers of x, but instead just keeping them implicitly in a tabular form in the same format as in the long division process, and just writing the coefficients. That's called "synthetic division" or some such. But it's just the same idea, just a little less writing.

-- Cheers, --Mr. d.