Factorizing/solving for x in simple quadratic

Question

I have a quadratic equation:

$$6x^2 - 19x + 10 = 0$$

I don't understand how to factorize it and thus solve for $x$ because of the $6$ out the front. If someone could show me in steps how to achieve the answer of $x=\frac23$, $x=\frac52$, it would be greatly appreciated. Hopefully from this i will be able to do the other questions in a similar format such as

$$10x^2 -9x +2 = 0$$

Answer 1

$$6x^2-19x+10=0$$ $$6x^2-15x-4x+10=0$$ $$3x(2x-5)-2(2x-5)=0$$ $$(3x-2)(2x-5)=0$$

Hope it helps?

For your second format, you can write it as:

$$10x^2-5x-4x+2=0$$ and proceed

Answer 2

We can use the Completing the Square Method for this:

\begin{align}6x^2-19x+10&=0\\ x^2-\frac{19}{6}x+\frac{10}{6}&=0\\ \left(x-\frac{\frac{19}{6}}{2}\right)^2-\left(\frac{\frac{19}{6}}{2}\right)^2+\frac{10}{6}&=0\\ \left(x-\frac{19}{12}\right)^2-\frac{361}{144}+\frac{10}{6}&=0\\ \left(x-\frac{19}{12}\right)^2-\frac{121}{144}&=0\\ \left(x-\frac{19}{12}\right)^2&=\frac{121}{144}\\ x-\frac{19}{12}&=\pm\sqrt{\frac{121}{144}}\\ x-\frac{19}{12}&=\pm\frac{11}{12}\\ x&=\frac{19}{12}\pm\frac{11}{12} \end{align}

And therefore we have \begin{align}x&=\frac{19}{12}+\frac{11}{12}\\ &=\frac 52\end{align}

or \begin{align}x&=\frac{19}{12}-\frac{11}{12}\\ &=\frac 23\end{align}

And so we can say that \begin{align}6x^2-19x+10&=\left(x-\frac 52\right)\left(x-\frac 23\right)\\ &=(2x-5)(3x-2)\end{align}

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Completing the Square

If we have an equation of the form $x^2+ax+b$, we can rewrite it as $$\left(x+\frac a2\right)^2-\left(\frac a2\right)^2+b$$

Answer 3

\begin{align}6x^2−19x+10=0\\ 6x^2−15x−4x+10=0\\ 3x(2x−5)−2(2x−5)=0\\ (3x−2)(2x−5)=0\end{align}

Now for the above equation to be equal to zero, either $(3x-2)$ needs to be equal to zero or $(2x-5)$ needs to be equal to zero.They both can also be equal to zero. So,

$3x-2=0$
$x=2/3$

Similarly, $x$ can also be equal to $5/2$.

For your second problem, you can rewrite it as

\begin{align}10x^2−5x−4x+2=0\end{align}

Can you take it from here?

Answer 4

A nice, rigorous way to think about it is:

$(px+q)(rx+s)≡ prx^2 + psx + qrx + qs $ (the general case)

For your quadratic, take the RHS of the equation and match the $x^2$, $x$ and "constant" terms:

$prx^2+psx+qrx+qs≡6x^2-19x+10$

$ \implies prx^2≡6x^2 \implies pr=6$

and

$ \implies psx+qrx≡-19x \implies ps + qr = -19$

and

$ \implies qs=10$

Now, although you could solve these three equations simultaneously to work out p, q r and s, it's much easier to try and find the values for these variables as the possibilities are limited. Assuming a nice factorisation (most of the questions are designed in this way), you will find the values are:

$p=3, q= -2, r=2, s=-5$ (you could swap around BOTH $p$ & $r$ AND $q$ & $s$ but it should be quite obvious as to why this should lead to the same factorisation)

So the factorisation is:

$(3x-2)(2x-5)=0 \implies x_1=\frac 23$ and $x_2=\frac52$

Some useful notes for these quadratics:

1. Note that when the $x^2$ coefficient is $1$ (maths jargon for saying that there is no number in front of $x^2$), $p$ and $s$ are both equal to $1$, which eliminates the first equation and makes the other two expressions much easier to solve, which is why this was never explained to you for these simpler types of quadratics.

2. As you do more and more of these, the process will become far quicker and you certainly won't need to go through this identity business (this is just to show you the theory behind the factorisation). An easier way to remember this is that the product of the constant terms in the factorisation must equal the constant term in the expanded quadratic, and that the [product of the "outside" terms] + [product of "inside" terms] must equal the middle $x$ term in the expanded quadratic (bear in mind that either of these products can be negative).

3. For quadratics with $x^2$ coefficient equal to 1 (as discussed above), it's far easier: you simply need to find two numbers $a$ and $b$ whose product equals the constant term and whose sum equals the co-efficient of the middle term, as the factorisation will be of the form $(x+a)(x+b)$ (again note that either or both of $a$ and $b$ can be negative).

4. Obviously not all quadratics can be factored ($x^2-x-1=0$ is a famous example, whose solutions are the golden ratios for positive and negative numbers respectively). In such cases, it should become obvious after a short while that the equation cannot be factored, in which case you should use the quadratic formula (alternatively complete the square as shown above) to solve for $x$.

Of course, this is a lot to take in, but hopefully as you become more experienced with quadratics, these rules will become ingrained and you won't need to think about these ideas at all.