Question

(y + 1) (y+4) (y-5) (y - 8) + 50.​

Answer 1

$\large\text{ \boxed{\bold{[Topic]}} }$

Factorization

Let's start solving our problem. To make a substitution, we need iterating terms. The idea to factorize this question is grouping.

$\large\text{ \boxed{\bold{[Step\ 1.]}} }$

According to the multiplicative property, the result of a product doesn't rely on the order of operations. So, let's make some common terms using products.

$\large\text{ \cdots\longrightarrow(y+4)(y-8)=\underline{y^{2}-4y}-32 }$

$\large\text{ \cdots\longrightarrow(y+1)(y-5)=\underline{y^{2}-4y}-5 }$

As we see here, we have a common part $\large\text{ \underline{y^{2}-4y} }$.

Let's choose any variable. Here, I choose $\larget$.

Let $\larget=y^{2}-4y$. Then what do we observe?

$\large\cdots\longrightarrow(t-32)(t-5)+50$

This is another quadratic polynomial for $\larget$, obtained by substitution.

$\large\text{ \boxed{\bold{[Step\ 2.]}} }$

After some calculation steps, we have -

$\large\cdots\longrightarrow t^{2}-37t+210$

After factorization, -

$\large\cdots\longrightarrow (t-30)(t-7)$

After substitution, -

$\large\cdots\longrightarrow (y^{2}-4y-30)(y^{2}-4y-7)$

And, this polynomial is no longer factorizable.

$\large\text{ \boxed{\bold{[Final\ answer]}} }$

So, we finish our answer.

$\large\text{ \cdots\longrightarrow\boxed{\bold{(y^{2}+4y-30)(y^{2}-4y-7)}} }$

$\large\text{ \boxed{\bold{[Learn\ more]}} }$

Quadratic polynomials

Factorization over integers

Any factorizable quadratic polynomial with integer coefficients can be verified by discriminant.

Let's say if we can factorize the quadratic factor $\largey^{2}+4y-30$.

$\large\cdots\longrightarrow D/4=2^{2}-1\times(-30)=34$

As we see here, 34 is not a perfect square. So, reducing is impossible.

Then, what about $\largey^{2}+4y-7$?

$\large\cdots\longrightarrow D/4=2^{2}-1\times(-7)=11$

11 is not a perfect square as well. So, reducing is impossible.

Quarter discriminant and the formula

Any quadratic polynomial with an even coefficient on the linear term is $\largeax^{2}-2b'x+c$.

Applying discriminant gives-

$\large\cdots\longrightarrow D=4b'^{2}-4ca=4(b'^{2}-ca).$

Dividing equation by 4, we get -

$\large\text{ \cdots\longrightarrow\boxed{D/4=b'^{2}-ca.} }$

Also, applying quadratic formula gives -

$\large\text{ \cdots\longrightarrow\boxed{x=\dfrac{-b'\pm\sqrt{D/4}}{a}.} }$

These two can save an amount of time in tests.