Question

is a zero of $(49x^2-1)+(1+7x{)}^2$ ?

• A. $\frac{1}{7}$
• B. $\frac{-1}{7}$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $\frac{-1}{7}$

$\text{Given},(49x^2-1)+(1+7x{)}^2=((7x{)}^2-1^2)+(1+7x{)}^2=(7x-1\left)\right(7x+1)+(1+7x\left)\right(1+7x)=(1+7x\left)\right(7x-1+1+7x\left)\right[\text{Taking}(1+7x)\text{common}]=(1+7x\left)\right(14x)$

$\text{To find the zeroes of a polynomial}\text{expression, we must equate it to zero}.(1+7x)\left(14x\right)=0\Rightarrow (1+7x)=0;\left(14x\right)=0\therefore x=\frac{-1}{7},0\text{Zeroes are}\frac{-1}{7}\&0.$