How do you factor #12y^2-7y+1 #?

Question

Avery Alexander · Accepted Answer

#color(blue)((3y-1)(4y-1) # is the factorised form of the expression.

#12y^2−7y+1 # We can Split the Middle Term of this expression to factorise it. In this technique, if we have to factorise an expression like #ay^2 + by + c#, we need to think of 2 numbers such that: #N_1*N_2 = a*c = 12*1 = 12# and, #N_1 +N_2 = b = -7# After trying out a few numbers we get #N_1 = -3# and #N_2 =-4# #-3*-4 = 12#, and #-3+(-4)= -7# #12y^2−color(blue)(7y)+1 =12y^2−color(blue)(3y -4y)+1 # #= 3y(4y-1) -1(4y-1) # #color(blue)((3y-1)(4y-1) # is the factorised form of the expression.

Savannah Anderson · Answer

undefined To factor $12y^2 - 7y + 1$, we need to find two numbers that multiply to $12 \times 1 = 12$ (the coefficient of $y^2$ multiplied by the constant term) and add up to $-7$ (the coefficient of $y$). These numbers are $-3$ and $-4$.

So, we can rewrite the expression as $12y^2 - 3y - 4y + 1$.

Then, we group the terms: $(12y^2 - 3y) + (-4y + 1)$.

Next, we factor out the greatest common factor (GCF) from each group: $3y(4y - 1) - 1(4y - 1)$.

Now, we can see that both groups share the common factor $4y - 1$.

So, the factored expression is $(3y - 1)(4y - 1)$.