Is it possible to factor y=9x^2 - 21x + 10 ? If so, what are the factors?

Jul 31, 2016

$y = \left(3 x - 2\right) \left(3 x - 5\right)$

Explanation:

Find the factors of 9 and 10 which combine and add up to 21. The signs will be the same, they are both minus.

One clue is to look at the size of the coefficient of the middle term.

$21$ is quite small compared to $9 \mathmr{and} 10$ which multiply to give $90$.

That is an indication that we are using the "middle factors"

Factors of 9: 1, 3 9. Factors of 10: 1,2,5,10.

Try the cross-products of $3 \mathmr{and} 3$ and $2 \mathmr{and} 5$ first.

$\text{ 3 2 } \Rightarrow 3 \times 2 = 6$
$\text{ 3 5 "rArr 3 xx5 = 15 " } 6 + 15 = 21$

The numbers in the top row are in one bracket.
The numbers in the bottom row are in one bracket.

$\left(3 x - 2\right) \left(3 x - 5\right)$

Multiply out for yourself to see how the numbers work.

Jul 31, 2016

$y = \left(3 x - 5\right) \left(3 x - 2\right)$

Explanation:

To determine that a given quadratic polynomial like $a {x}^{2} + b x + c$

can be factorised or not in $\mathbb{R}$, we can use the following test :-

The pol. can have factors in $\mathbb{R} \iff \Delta = {b}^{2} - 4 a c \ge 0$.

In our case, $\Delta = {\left(- 21\right)}^{2} - 4 \cdot 9 \cdot 10 = 441 - 360 = 81 \ge 0$.

Therefore, the poly. $y$ can be factorised in $\mathbb{R}$.

Observe that, $9 \times 10 = 90$, so to factorise the given expression, we

should find $2$ factors of $90$, such that the sum of these factors

be $21$=middle term.

We find, $15 \times 6 = 90 , \mathmr{and} , 15 + 6 = 21$.

Hence, $y = 9 {x}^{2} - 21 x + 10 = 9 {x}^{2} - 15 x - 6 x + 10$

$= 3 x \left(3 x - 5\right) - 2 \left(3 x - 5\right) = \left(3 x - 5\right) \left(3 x - 2\right)$

Enjoy Maths.!