Question

How do you factor \[4{{x}^{2}}-15x+9\]?

Answer 1

Hint: In order to find the solution of the given question that is to factor the quadratic expression \[4{{x}^{2}}-15x+9\] use the sum-product pattern or we can say using splitting the middle term that is to find factors of \[4\] and \[9\] which combine and add to give \[15\]. Then common factors from two pairs and rewrite them in factored form.

Complete step by step solution:
According to the question, given expression in the question is as follows:
\[4{{x}^{2}}-15x+9\]
We will factorise the above expression by using the splitting the middle term method or we can sum-product pattern, we will have to find factors of \[4\] and \[9\] which combine and add [because of positive \[9\].] to give \[15\].
The signs in the brackets will be the same (because of positive \[9\]) and they will both be negative (because of negative \[15\].)
So, find the factors of \[4\] and \[9\] then cross multiply, we will have:
Factors of \[4\]: \[4\] and \[1\].
Factors of \[9\]: \[3\] and \[3\].
Here we see that \[1\times 3=3\] and \[4\times 3=12\]. And \[3+12=15\]. This means we can split \[-15x\] into \[-3x-12x\].
\[\Rightarrow 4{{x}^{2}}-3x-12x+9\]
After the above step take the terms in common from the two pairs, we will have:
\[\Rightarrow x\left( 4x-3 \right)-3\left( 4x-3 \right)\]
Now rewrite the above expression in the factored-form, we get the final answer as:
\[\Rightarrow \left( 4x-3 \right)\left( x-3 \right)\]
Therefore, the factor-form of the given expression \[4{{x}^{2}}-15x+9\] is \[\left( 4x-3 \right)\left( x-3 \right)\].

Note: There is a clue in the fact that \[15\] is odd. An odd number is formed from an odd + even. This tells us that \[4\] didn’t split as \[2\times 2\], because then both products would be even. Even + even = even. Therefore \[4\] has to be used as \[4\times 1\].