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How do you factor,
$21{{x}^{2}}+55x+14=0?$

seo-qna
Last updated date: 26th Feb 2024
Total views: 339.9k
Views today: 6.39k
IVSAT 2024
Answer
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Hint: As, quadratic equation are those equation in which maximum power of variable is $2.$
It is in form of $a{{x}^{2}}+bx+c=0$
Where, $a$ is co-efficient of $x$
$b$ is co-efficient of $x$
And $c$ is constant.
Here, value of $a=1,b=-11,c=24$
For factoring, we need two numbers which when added or subtracted together to form the product of $'a'\And 'c'$
Like, ${{x}^{2}}+2x-2=0$
$a=1,b=2,c=2$
So, $ac=\left( 1\times 2 \right)=2$
So, factor or number will be $2$ and $1$
Hence, we can write it as,
${{x}^{2}}+x-2x-2=0$
Apply this concept to factorize the given expression.

Complete step by step solution:
As per data given in the question,
We have,
$21{{x}^{2}}+55x+14=0$
As we know that,
General form of a quadratic equation is,
$a{{x}^{2}}+bx+c=0$
Where, $a\And b$ are coefficients of ${{x}^{2}}$ and $x$ respectively.
And $c$ is any constant.
So, after comparing the given equation with the general equation of the quadratic equation.
We will get,
$a=21,b=55,c=14$
So, value of product of $a$ and $c$ will be $=\left( a\times c \right)=\left( 21\times 14 \right)=294$
As for solving the above equation,
We need pairs of such numbers which when multiplied result to $294$
And when added/subtracted together gives $55.$
So,
Factorizing $294,$we will get,
$294=2,147\And \left( -2,-147 \right)$
$294=3,98\And \left( -3,-98 \right)$
$294=6,49\And \left( -6,-49 \right)$
$294=7,42\And \left( -7,-42 \right)$
Hence, from above factorization,
We can conclude that,
Only pairs of numbers $\left( 6,49 \right)$ and $\left( -6,-49 \right)$ can gives a value of $55.$
As, we need a positive value of $'b'$
So, we will consider number $6,49$ only
Hence,
Above equation becomes,
$21{{x}^{2}}+55x+14=0$
$\Rightarrow 21{{x}^{2}}+6x+49+14=0$
Taking $3$ as common from first two digits and $7$ as common from last two digits,
We will get,
$21{{x}^{2}}+6x+49x+14=0$
$3x\left( 7x+2 \right)+7\left( 7x+2 \right)=0$
$\left( 3x+7 \right)\left( 7x+2 \right)=0$

Hence, In factorization form,
$21{{x}^{2}}+55x+14=0$ can be written as $\left( 3x+7 \right)\left( 7x+2 \right)$


Additional Information:
Let if we have to factorize,
${{x}^{2}}+3x+2=0$
Here, sign of all co-efficient are $+ve$
So,
$ac=2\times 1=2$
Hence, two number can be, $\left( 2,1 \right)$ and $\left( -1,-2 \right)$
But due to positive sign of $''b''$ we will consider number as $\left( 2,1 \right)$
$+\times +=+$
$+\times -=-$
$-\times +=-$
$-\times -=+$

Note: Let we have to factorize the equation ${{x}^{2}}-11x+24=0,$ As, in the given equation,
${{x}^{2}}-11x+24=0$
We will not consider $\left( 8,3 \right)$ as pairs of numbers, as here $''b''$ is negative and $''c''$ is positive, so we need
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