Factoring Polynomial using Quadratic EquationI'm trying to use the quadratic equation (QE) to factor a degree 2 polynomial into the format: ( x + a ) 2 , where a is any real number.This works great for equations like:(1) x 2 + 2 x + 1The QE gives roots x = { − 1 , − 1 }, and ( x + 1 ) 2 = x 2 + 2 x + 1.This approach fails for other polynomials such as:(2) 9 x 2 + 36 x + 36The roots found with the QE are x = { − 2 , − 2 }. But ( x + 2 ) 2 ≠ 9 x + 36 x + 36.9 is the GCF from the coefficients of (2), so (2) can be rewritten as: 9 × ( x 2 + 4 x + 4 )and by using the QE on the second term, this equals: 9 × ( x + 2 ) 2 . And 9 × ( x + 2 ) 2 = 9 x + 36 x + 36 (2), so this seems to be a better approach.So my question is: must the input to QF be a polynomial that has coefficients with G C F = 1, for the result to give roots that can be used to reconstruct the original function?

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Factoring Polynomial using Quadratic EquationI&#039;m trying to use the quadratic equation (QE) to factor a degree 2 polynomial into the format:   (  x  +  a      )    2  , where a is any real number.This works great for equations like:(1)       x    2    +  2  x  +  1The QE gives roots   x  =  {  −  1  ,  −  1  }, and   (  x  +  1      )    2    =      x    2    +  2  x  +  1.This approach fails for other polynomials such as:(2)   9      x    2    +  36  x  +  36The roots found with the QE are   x  =  {  −  2  ,  −  2  }. But   (  x  +  2      )    2    ≠  9  x  +  36  x  +  36.9 is the GCF from the coefficients of (2), so (2) can be rewritten as:  9  ×  (      x    2    +  4  x  +  4  )and by using the QE on the second term, this equals:  9  ×  (  x  +  2      )    2  . And   9  ×  (  x  +  2      )    2    =  9  x  +  36  x  +  36 (2), so this seems to be a better approach.So my question is: must the input to QF be a polynomial that has coefficients with   G  C  F  =  1, for the result to give roots that can be used to reconstruct the original function?

Arely Davila · Accepted Answer

Step 1Look at the expansion  (  x  +  a      )    2    =      x    2    +  2  a  x  +      a    2    .Notice that the coefficient of       x    2   is always 1. Thus, if a quadratic polynomial can be written in this form, it must have a leading coefficient of 1. In general, if you have a quadratic polynomial with leading coefficient c and it has a double root at -a, then the polynomial can be written as   c  (  x  +  a      )    2  . So, you don&#039;t want to be looking at the GCF, but rather at the leading coefficient. (Although these will be equal in any polynomial with positive leading coefficient and a double root)Step 2More generally, if you have a polynomial P(x) of degree d with leading coefficient c and roots (enumerated with multiplicity)       r    1    ,        r    2    ,  …  ,      r    d  , one gets  P  (  x  )  =  c  (  x  −      r    1    )  (  x  −      r    2    )  ⋯  (  x  −      r    d    )  .That is, the leading coefficient can always be pulled out of the factorized form. (Note that this doesn&#039;t work if you don&#039;t have d roots - if you&#039;re working in the real numbers and not the complex numbers, some polynomials can&#039;t be factored)

Kareem Mejia · Accepted Answer

Step 1Yes. As you have taken the format   (  x  +  a      )    2   which have coefficient of x as 1 so this format will work only for coefficient as 1 of a polynomial. As your other example has 9 as coefficient so you will have to take 3 as coefficient of x in the format   (  x  −  a      )    2  .Step 2As x will be 3x so a must be 3a. This corresponding change in format will give it new look as   (  3  x  −  3  a      )    2  . And if you put   (  3  x  −  3  ×  2      )    2   it will give answer   9      x    2    +  36  x  +  36.

Factoring Polynomial using Quadratic Equation. I'm trying to use the quadratic equation (QE) to factor a degree 2 polynomial into the format: (x+a)^2, where a is any real number. This works great for equations like: (1) x^2+2x+1.

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