
How do you factor ${{x}^{2}}+3x-4?$
Answer
545.4k+ views
Hint: Given polynomial is a second-degree polynomial and is also known as quadratic polynomial and the standard form of the second-degree polynomial is $A{{x}^{2}}+Bx+c$.
To the factor, a polynomial means to break the expression $(A{{x}^{2}}+Bx+c)$ down into smaller expressions that are multiplied together like as $(x+k)\times (x+h)$ here $k$ and $h$ are the constant.
If we multiply both factors $k$ and $h$ then they must be equal to constant $C$ and if we sum both factors they must be equal to the coefficient of the middle term.
$\Rightarrow k+h=C$ and
\[\Rightarrow k\times h=B\]
where $A$ is the coefficient of the term${{x}^{2}}$, $B$ is the coefficient of the term $x$, and $C$ is constant.
Complete step by step solution:
The given factor is ${{x}^{2}}+3x-4$.
In the above polynomial $A=1,B=3$and $C=-4$.
To factorize this polynomial find two factors $-4$ whose sum will be equal to the coefficient of the middle term of the above polynomial which is $3$
i.e $-1+4=3$
Here, $-1$ and $4$ are the factors of -4 whose sum will be equal to 3.
Now, rewrite the coefficient of the middle term using two factors $-1$ and $4$
$\Rightarrow {{x}^{2}}+(-1+4)x-4$
$\Rightarrow {{x}^{2}}-x+4x-4$
$\Rightarrow x(x-1)+4(x-1)$
$\Rightarrow (x-1)\times (x+4)$
Thus, $(x-1)$ and $(x+4)$ are desired factors of the ${{x}^{2}}+3x-4$.
Note: The second-degree polynomials are known as quadratic polynomials and it becomes quadratic when it will be equal to zero.
The shape of the quadratic polynomials is always parabolic.
The roots of quadratic polynomials are two real, zero, or one.
To the factor, a polynomial means to break the expression $(A{{x}^{2}}+Bx+c)$ down into smaller expressions that are multiplied together like as $(x+k)\times (x+h)$ here $k$ and $h$ are the constant.
If we multiply both factors $k$ and $h$ then they must be equal to constant $C$ and if we sum both factors they must be equal to the coefficient of the middle term.
$\Rightarrow k+h=C$ and
\[\Rightarrow k\times h=B\]
where $A$ is the coefficient of the term${{x}^{2}}$, $B$ is the coefficient of the term $x$, and $C$ is constant.
Complete step by step solution:
The given factor is ${{x}^{2}}+3x-4$.
In the above polynomial $A=1,B=3$and $C=-4$.
To factorize this polynomial find two factors $-4$ whose sum will be equal to the coefficient of the middle term of the above polynomial which is $3$
i.e $-1+4=3$
Here, $-1$ and $4$ are the factors of -4 whose sum will be equal to 3.
Now, rewrite the coefficient of the middle term using two factors $-1$ and $4$
$\Rightarrow {{x}^{2}}+(-1+4)x-4$
$\Rightarrow {{x}^{2}}-x+4x-4$
$\Rightarrow x(x-1)+4(x-1)$
$\Rightarrow (x-1)\times (x+4)$
Thus, $(x-1)$ and $(x+4)$ are desired factors of the ${{x}^{2}}+3x-4$.
Note: The second-degree polynomials are known as quadratic polynomials and it becomes quadratic when it will be equal to zero.
The shape of the quadratic polynomials is always parabolic.
The roots of quadratic polynomials are two real, zero, or one.
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