Please sketch and factorize the quadratic 3x^2+10x+3.

The coefficient - which is a posh word for 'the bit in front' - of x^2 is in this case not 1. Thus, the quadratic formula must necessarily be used. All quadratics come in the form of ax^2+bx+c. Here, a, b, and c, respectively are 3, 10, and 3. When placed within the formula of x=(-b+or-(b^2-4ac)^1/2)/2a the answers generated are x=(-10+8)/6 and x=(-10-8)/6 Hence, x=-1/3 and -3. These are the x coordinates at which the curve will hit the x-axis (where y=0). The curve will necessarily be a U shape because 'a' is positive (it's 3) in this case. Otherwise it would be an N shape (if you'll excuse the font). In order to factorize this, simply allow each equation to equal and zero and multiply away any denominators. x=-1/3 becomes x+1/3=0 which becomes 3x+1=0. Likewise, x=-3 becomes x+3=0. The factorized equation is therefore (3x+1)(x+3). Occasionally, when dealing with complex quadratics as these (where a does not equal 1), a bracket may need to be multiplied by an integer (whole, positive number) in order to have the two brackets multiply out to form the original equation. For example, 6x^2+20x+6 will have the same solutions for x as the above equation as it is simply the above equation multiplied by an integer value, in this case 2. Hence, for that equation, upon finding x to be -1/3 and -3, one could not simply eradicate the denominators through multiplication and rearrange it to equal zero but must multiply by 2. Whether there is a need to do this and by what integer will be obvious should one be on the look out for it.