Really need help for this question

Question

a) Factorise 3x^2+19x+20. 

b) Hence, Factorise 3(y-2)+19(y-2)+20

Answer 1

Okay, Amy,

 

Let's first look at factoring a quadratic abstractly so you can see what we are after:

 

We want a product of two monomials: (ax + b)(cx + d) = acx² + (ad + bc)x + bd

 

We can see that ac = 3, ad + bd = 19, and bd =20.

 

So, we have to factor ac = 3 = 3*1 and bd = 20 = 4*5 = 2² * 5.

 

Since there is only one way to factor ac, we can write the monomials as (3x + b)(1x +d) = 3x² + (3d + 1b)x +bd.

 

This means that 3d + b = 19 and that we might guess what b and d, the factors in bd = 20, are by trying various combinations of 4, 5, 2, and 10, the possible factors in 20 = bd.

 

So, let's try d = 2 and see (solve for) what b should be: 3*2 + b = 19 => b = 13, which is not a factor of 20.

 

So, let's try d = 4 and see (solve for) what b should be: 3*4 + b = 19 => b = 7, again not a factor of 20.

 

You should expect to try all the possible factors. One will work to make 3d + b = 19, but it might not be the first and it might be the last.

 

So, let's try d = 5 and see (solve for) what b should be: 3*5 + b = 19 => b = 4. That's it!

 

Finally, we have (3x + 4)(x + 5) = 3x² + (3*5 + 4*1)x + 4*5 = 3x² + 19x +20.

 

Notice that this solution was not too much work because the coefficient of x² had only two factors, 3 and 1. It is a good tactic to start with whichever the coefficient of x², ac = 3, or the constant coefficient, bd =20, has fewer factors. If neither has only two factors, expect to have to try many more combinations before you find the correct one.

 

Part b:

 

Because we know that 3x² + 19x + 20 = (3x + 4)(x + 5), we can substitute y-2 for x on both sides to get:

 

3(y - 2)² + 19(y - 2) + 20 = [3(y - 2) + 4][(y - 2) + 5] = (3y - 2)(y + 3).

 

Michael.