Question

X^2-11x-4x do it by middle term spilliting

Answer 1

Correct Question

• X²-11x+30 do it by middle term spilliting

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

• A polynomial
• X²-11x+30

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

• Factors of the polynomial

.

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

Method-1

• X²-11x+30

• we have to spilt the middle term in such a way that the product become 30 and sum become -11x

X²-11X+30

=>X²-5x-6x+30

=>x(x-5)-6(x-5)

=>(x-5) (x-6)

=>x=5,4

Method -2

By quadratic formula

Here

a=1

b=-11

C=30

$\tt{X=\dfrac{-b\pm{\sqrt{b^2-4ac}}}{2a{}$

¶utting values

$\tt\rightarrow{X=\dfrac{-b\pm{\sqrt{b^2-4ac}}}{2a}}$

$\tt\rightarrow{X=\dfrac{11\pm{\sqrt{-11^2-4\times{1}\times{30}}}}{2}}$

$\tt\rightarrow{X=\dfrac{11\pm{\sqrt{121-120}}}{2}}$

$\tt\rightarrow{X=\dfrac{11\pm{\sqrt{1}}}{2}}$

Taking X as +

$\tt\rightarrow{X=\dfrac{11+{\sqrt{1}}}{2}}$

=>{11+1}/2

=>6

Taking X as +

$\tt\rightarrow{X=\dfrac{11-{\sqrt{1}}}{2}}$

=>{11-1}/2

=>5

Additional Information

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

Then,

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

&

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

☞

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{11}{1}$

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{Cofficient\:of\:X}{Cofficient\:of\:x^2}=$

&

$\rightarrow\tt\alpha{\times}\beta{=}\dfrac{30}{1}$

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

Hence,relation verified

Answer 2

Hello,

$\longmapsto\tt{{x}^{2}-11x-30}\\ \\\longmapsto\tt{{x}^{2}-(5x-6x)+30}\\ \\ \longmapsto\tt{{x}^{2}-5x+6x+30}\\ \\ \longmapsto\tt{x(x-5)-6(x-5)}\\ \\ \longmapsto\tt{(x-6)(x-5)}$