i need help in factorization of algebric expressions?

a sum:

xsquare- (p+q)x+pq

8 Answers

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  • Som™
    Lv 6
    1 decade ago
    Favorite Answer

    x^2- (p+q)x+pq

    To factorize this equation :

    1. You need to factorize/break the number pq into its products.

    2. Using the factors obtained by above step ,by suitable addition or subtraction obtain the number -(p+q)

    For this sum

    factor of pq are p and q or

    -p and -q

    Adding the latter we get:

    (-p)+(-q)

    = -(p+q)

    Thus we get

    x^2- px-qx+pq

    =x(x-p)-q(x-p)

    =(x-p)(x-q)

    Source(s): Hope you get it
  • 1 decade ago

    That's how you factor a quadratic that has lead coeffeicnt 1. Find two numbers (p and q) that you can multiply together to get the last (constant) number, and add togeterh to get the middle (linear coeffieicnt) number. Note: the - in between the first and second terms does not necessarily mean that you can't have a + there in your example, it just means that the sum (p+q) may be positive or negative)

    So then your factors are (x-p)(x-q). Multiply it out by FOIL and you get x^2 - px - qx + pq = x^2 - (p+q)x + pq

    Often people have a hard time learning to factor quadratics, then when they see this pattern they have an aha! moment. Hope that comes for you.

    Next is to learn to factor quadratics that have a lead coeffieicnt other than zero. Most texts and teachers can offer little more than brute force trial and error, and many students give up on this type of question. But there is a cool trick for finding the factors of a quadratic like this. It's kind of hard to write up in this media, though. I did it once, I'll try and find it and put it on my 360 page for future reference.

  • 3 years ago

    answer: a million. (a) factorise 4x(sq.) -25 =(2x-5)(2x+5) * basically use the go technique. (b) (i) U basically equate the factorize in the past with 0. like under: (2x-5)(2x+5) = 0, then u will have been given the respond as x =+/- 2.5 (ii) (2x-5)(2x+5) < 0 . it quite is potential that y is comparable to detrimental fee.. Then, u'll have been given 2 solutions it relatively is x<-2.5 and x<2.5 so, to make the value x fulfill the the two one in each and every of condition u basically take x is below 2.5 through fact the respond. staggering? u can verify via calculator. thank you..

  • 1 decade ago

    x^2-(p+q)x+pq

    =x^2-px-qx+pq

    =x(x-p)-q(x-p)

    =(x-p)(x-q) Ans.

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  • 1 decade ago

    x^2 - (p+q)x + pq

    = x^2 -px -qx + pq

    = x^2 - qx - px + pq

    = x(x - q) - p(x - q)

    = (x - p) (x - p)

  • 1 decade ago

    x^2-x[p+q]+pq

    =x^2-px-qx+pq

    =x[x-p]-q[x-p]

    =[x-q][x-p]

  • 1 decade ago

    xsquare-px-qx+pq

    =x(x-p)-q(x-p)

    =(x-p)(x-q)

    Therefore, xsquare-(p+q)x+pq= (x-p)(x-q)

  • Anonymous
    1 decade ago

    =(x-p)(x-q)

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