Question

Factorise $15y^2-8y+1$ by using the factor theorem

• A. $(3y-1)(5y-1)$
• B. $(3y+1)(5y-1)$
• C. $(3y-1)(5y+1)$
• D. $(3y+1)(5y+1)$

Answer 1

Answer: (A)

$(3y-1)(5y-1)$

Let $p\left(y\right)=15y^2-8y+1=15\times \{y^2-\frac{8}{15}y+\frac{1}{15}\}$

$=15\times q\left(y\right)$

where $q\left(y\right)=y^2-\frac{8}{15}y+\frac{1}{15}$

We have made the coefficient of the leading term of the polynomial q(y) equal to 1. The constant term of the quadratic polynomial q(y) is $\frac{1}{15}$. Some of the factors of the number $\frac{1}{15}$ are $\pm 1,\pm \frac{1}{3},\pm \frac{1}{5},\pm \frac{1}{15}$

We find that
$q\left(\frac{1}{3}\right)={\left(\frac{1}{3}\right)}^2-\frac{8}{15}\left(\frac{1}{3}\right)+\frac{1}{15}$

$=\frac{1}{9}-\frac{8}{45}+\frac{1}{15}$

$=\frac{5-8+3}{45}=0$

and $q\left(\frac{1}{5}\right)={\left(\frac{1}{5}\right)}^2-\frac{8}{15}\left(\frac{1}{5}\right)+\frac{1}{15}$

$=\frac{1}{25}-\frac{8}{75}+\frac{1}{15}$

$=\frac{3-8+5}{75}=0$

Thus, $\frac{1}{3}$ and $\frac{1}{5}$ are zeros of q(y). It implies that

$(y-\frac{1}{3})$ and $(y-\frac{1}{5})$ are two factors of the polynomial $q\left(y\right)=y^2-\frac{8}{15}y+\frac{1}{15}$ (By factor theorem).

i.e., $y^2-\frac{8}{15}+\frac{1}{15}=(y-\frac{1}{3})(y-\frac{1}{5})$

$\Rightarrow 15y^2-8y+1=15\times \left\{(y-\frac{1}{3})(y-\frac{1}{5})\right\}$

$=15\times \{\frac{(3y-1)}{3}\times \frac{(5y-1)}{5}\}$

$=(3y-1)(5y-1)$

Therefore, the polynomial $15y^2-8y+1$ factorised into two linear factors as $(3y-1)(5y-1)$