Question

Find the zeros
v²+4√3v-15 verification

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• A polynomial
• V²+4√3v-15

${\sf{\green{\underline{\large{To\:Find}}}}}$

• Factors of the polynomial
• Relationship between cofficient

${\sf{\pink{\underline{\Large{Explanation}}}}}$

V²+4√3v-15

we have to spilt the middle term in such a way that the product become -15and sum become 4√3

V²+4√3v-15

=>V²+5√3V-√3V-15

=>V(v+5√3)-√3(v+5√3)

=>(v-√3) (v+5√3)

Additional Information

Let the zeroes of the polynomial be$\tt\alpha{and}\beta$

Then,

$\rightarrow\tt\alpha{+}\beta{=}\frac{-b}{a}$

&

$\rightarrow\tt\alpha{\times}\beta{=}\frac{c}{a}$

Here,

a=1

b=4√3

C=-15

☞

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{4\sqrt{3}}{1}$

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{-(Cofficient\:of\:V)}{Cofficient\:of\:V^2}=$

&

$\rightarrow\tt\alpha{\times}\beta{=}\dfrac{-15}{1}$

$\rightarrow\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^2}}$

Hence,relation verified

Answer 2

Hello,

ANSWER : -

$\longmapsto\tt\bold{{v}^{2}+4\sqrt{3}v-15}$

→Now, Split the middle term in the way that sum becomes 4√3 and Product -15 ..

$\longmapsto\tt\bold{{v}^{2}+4\sqrt{3}v-15}\\ \\ \longmapsto\tt\bold{{v}^{2}+5\sqrt{3}v-\sqrt{3}v-15}\\ \\ \longmapsto\tt\bold{v(v+5\sqrt{3})-\sqrt{3}\:\:(v+5\sqrt{3})}\\ \\ \longmapsto\tt\bold{(v-\sqrt{3}\:\: (v+5\sqrt{3})}$